UC-NRLF 


$B    5?^    133 

INTRODUCTORY 


MODERN    GEOMETRY 


OF 


POINT,  RAY,  AND  CIRCLE 


PART  I 


BY 


W.    B.    SMITH,    Ph.D. 


^-~  ■■■      o 


Ij^ 


IN  MEMORIAM 
FLORIAN  CAJORI 


f/a^    C?/y-^^-^ 


The  price  of  Part  I.  is  75  cents,  and  of  the  co7?iplete 
hook  $1.10.  Copies  of  this  part,  if  returried  to  the  pub- 
lishers in  good  condition,  will  be  exchanged  for  the  complete 
book,  on  payment  of  the  difference  in  price. 


INTRODUCTORY  MODERN  GEOMETRY 


.^>f)m 


INTRODUCTORY 

MODERN  GEOMETRY 

OF 

POINT,  RAY,  AND  CIRCLE 
Part  I 

BY 

WILLIAiVrBENJAMIN  SMITH,  A.M.,  Ph.D.  (Goett.) 

Professor  of  Mathematics  and  Astronomy 
University  of  the  State  of  Missouri 


aeltefleS  berool^rt  mit  trcue, 
^reunblid(>  aufgefagted  3leue. 

—  Goethe. 


Ncto  gorft 
MACMILLAN    &   CO. 

AND    LONDON 
1892 

All  rights  reserved 


Copyright,  1892, 
By  MACMILLAN  AND  CO. 


Typography  by  J.  S.  Cushing  &  Co.,  Boston,  U.S.A. 
Presswork  by  Berwick  &  Smith,  Boston,  U.S.A. 


GEOMETRY. 


INTRODUCTION. 

1 .  Geometry  is  the  Doctrine  of  Space. 

What  is  Space?  On  opening  our  eyes  we  see  objects 
around  us  in  endless  number  and  variety :  the  book  here, 
the  table  there,  the  tree  yonder.  This  vision  of  a  world 
outside  of  us  is  quite  involuntary  —  we  cannot  prevent  it, 
nor  modify  it  in  any  way ;  it  is  called  the  Intuition  (or  Per- 
ception or  Envisagement)  of  Space.  Two  objects  precisely 
alike,  as  two  copies  of  this  book,  so  as  to  be  indistinguishable 
in  every  other  respect,  yet  are  not  the  same,  because  they 
differ  in  place,  in  their  positions  in  Space  :  the  one  is  here, 
the  other  is  not  here,  but  there.  In  between  and  all  about 
these  objects  that  thus  differ  in  place,  there  lies  before  us 
an  apparently  unoccupied  region,  where  it  seems  that  noth- 
ing isy  but  where  anything  might  be.  We  may  imagine  or 
suppose  all  these  objects  to  vanish  or  to  fade  away,  but  we 
cannot  imagine  this  region,  either  where  they  were  or  where 
they  were  not,  to  vanish  or  to  change  in  any  way.  This 
region,  whether  occupied  or  unoccupied,  where  all  these 
objects  are  and  where  countless  others  might  be,  is  called 
Space. 

2.  There  are  certain  elementary  facts ^  that  is,  facts  that 
cannot  be  resolved  into  any  simpler  facts,  about  this  Space, 
and  these  deserve  special  notice. 


K7i«nf».*ii 


2  GEOMETRY. 

A.  Space  is  fixed,  permanent,  unchangeable.  The  objects 
in  Space,  called  bodies,  change  place,  or  may  be  imagined 
to  change  place,  in  all  sorts  of  ways,  without  in  the  least 
affecting  Space  itself.  Animals  move,  that  is,  change  their 
places,  hither  and  thither ;  clouds  form  and  transform  them- 
selves, drifting  before  the  wind,  or  dissolve,  disappearing 
altogether ;  the  stars  circle  eternally  about  the  pole  of  the 
heavens ;  sun,  moon,  and  planets  wander  round  among  the 
stars ;  but  the  blue  dome  of  the  sky,*  the  immeasurable 
expanse  in  which  all  these  motions  go  on,  remains  unmoved 
and  immovable,  as  a  whole  and  in  all  its  parts,  absolutely 
the  same  yesterday,  to-day,  and  forever. 

B.  Space  is  homoBoidal ;  i.e.  it  is  precisely  alike  through- 
out its  whole  extent.  Any  body  may  just  as  well  be  here, 
there,  or  yonder,  so  far  as  Space  is  concerned.  A  mere 
change  of  place  in  nowise  affects  the  Space  in  which  the 
change,  or  motion,  occurs. 

C.  Space  is  boundless.  It  has  no  beginning  and  no  end. 
We  may  imagine  a  piece  of  Space  cut  out  and  colored  (to 
distinguish  it  from  the  rest  of  Space)  ;  the  piece  will  be 
bounded,  but  Space  itself  will  remain  unbounded. 

N.  B.  When  we  say  that  Space  is  unbounded,  we  do  not 
mean  that  it  is  infinite.  Suppose  an  earthquake  to  sink  all 
the  land  beneath  the  level  of  the  sea,  and  suppose  this  latter 
at  rest ;  then  its  outside  would  be  unbounded,  without  begin- 
ning and  without  end,  —  a  fish  might  swim  about  on  it  in 
any  way  forever,  without  stop  or  stay  of  any  kind.  But  it 
would  7iot  be  infinite ;  there  would  be  exactly  so  many 
square  feet  of  it,  a  finite  number,  neither  more  nor  less. 
Likewise,  the  fact  that  bodies  may  and  do  move  about  in 
space  every  way  without  let  or  hindrance  of  any  kind  implies 

*  Appearing  blue  because  of  the  refraction  of  light  in  the  air. 


INTRODUCTION.  3 

that  Space  is  boundless,  but  by  no  means  that  it  is  infinite. 
For  all  we  know  there  may  be  just  so  many  cubic  feet  of 
Space ;  it  may  be  just  so  many  times  as  large  as  the  sun, 
neither  more  nor  less.  This  distinction  between  unbounded 
and  infinite,  first  clearly  drawn  by  Riemann,  is  fundamental. 

D.  Space  is  continuous.  There  are  no  gaps  nor  holes  in 
it,  where  it  would  be  impossible  for  a  body  to  be.  A  body 
may  move  about  in  Space  anywhere  and  everywhere,  ever 
so  much  or  ever  so  litUe.  Space  is  itself  simply  where  a 
body  may  be,  and  a  body  may  be  anywhere. 

E.  Space  is  triply  extended,  or  has  three  dimensions. 
This  important  fact  needs  careful  explication. 

In  telling  the  size  of  a  box  or  a  beam  we  find  it  necessary 
and  sufficient  to  tell  three  things  about  it :  its  length,  its 
breadth,  and  its  thickness.  These  are  called  its  dimensions  ; 
knowing  them,  we  know  the  size  completely.  But  to  tell  the 
size  of  a  ball  it  is  enough  to  tell  one  thing  about  it,  namely, 
its  diameter;  while  to  tell  the  size  of  a  chair  we  should 
have  to  tell  many  things  about  it,  and  we  should  be  puzzled 
to  say  what  was  its  length,  or  breadth,  or  thickness.  Never- 
theless, it  remains  true  that  Space  and  all  bodies  in  Space 
have  just  three  dimensions,  but  in  the  sense  now  to  be  made 
clear. 

We  learn  in  Geography  that,  in  order  to  tell  accurately 
where  a  place  is  on  the  outside  of  the  earth,  which  may 
conveniently  be  thought  as  a  level  sheet  of  water,  it  is 
necessary  and  sufficient  to  tell  two  things  about  it ;  namely, 
its  latitude  and  its  longitude.  Many  places  have  the  same 
latitude,  and  many  the  same  longitude ;  but  no  two  have 
the  same  latitude  and  the  same  longitude.  It  is  not  suffi- 
cient, however,  if  we  wish  to  tell  exactly  where  a  thing  is  in 
Space,  to  tell  two  things  about  it.  Thus,  at  this  moment 
the  bright  star  Jupiter  is  shining  exacdy  in  the  south ;  we 


4  GEOMETRY. 

also  know  its  altitude,  how  high  it  is  above  the  horizon  (this 
altitude  is  measured  angularly  —  a  term  to  be  explained 
hereafter,  but  with  which  we  have  no  present  concern) .  But 
the  knowledge  of  these  two  facts  merely  enables  us  to  poiiit 
towards  Jupiter ;  they  do  not  fix  his  place  definitely,  they 
do  not  say  how  far  away  he  is  :  we  should  point  towards 
him  the  same  way  whether  he  were  a  mile  or  a  million  of 
miles  distant.  Accordingly,  a  third  thing  must  be  known 
about  him,  in  order  to  know  precisely  where  he  is  ;  namely, 
his  distance  from  us.  But  when  this  third  thing  is  known, 
no  further  knowledge  about  his  place  is  either  necessary  or 
possible.  Once  more,  here  is  the  point  of  a  pin.  Where 
is  it  in  this  room  ?  It  is  five  feet  above  the  floor.  This  is 
not  enough,  however,  for  there  are  many  places  five  feet 
above  the  floor.  It  is  also  ten  feet  from  the  south  wall,  but 
there  are  yet  many  positions  five  feet  from  the  floor  and  ten 
feet  from  the  south  wall,  as  we  may  see  by  sHpping  a  cane 
five  feet  long  sharpened  to  a  point,  upright  on  the  floor, 
keeping  the  point  always  ten  feet  from  the  south  wall.  But 
as  it  is  thus  slipped  along,  the  point  of  the  cane  will  come 
to  the  point  of  the  pin  and  then  will  be  exactly  twelve  feet 
from  the  west  wall.  If  it  now  move  ever  so  little  either  way 
east  or  west,  it  will  no  longer  be  at  the  pin-point  and  no 
longer  twelve  feet  from  the  west  wall.  So  there  is  one,  and 
only  one,  point  that  is  five  feet  from  the  floor,  ten  feet  from 
the  south  wall,  and  twelve  feet  from  the  west  wall.  Hence 
it  is  seen  that  these  three  facts  fix  the  position  of  the  pin- 
point exactly.  A  fourth  statement,  as  that  the  point  is  nine 
from  the  ceiling,  will  either  be  superfluous,  if  the  ceiling  is 
fourteen  feet  high,  being  implied  in  what  is  already  said,  or 
else  incorrect,  if  the  ceiling  is  not  fourteen  feet  high,  contra- 
dicting what  is  already  said.  In  general,  with  respect  to  any 
position  in  Space  it  is  necessary  to  know  three  independent 


JXTROnUCTION.  5 

facts  (or  data),  and  it  is  impossible  to  know  any  more.  All 
other  knowledge  about  the  position  is  involved  in  this  knowl- 
edge, which  is  necessary  and  sufficient  to  enable  us  to  an- 
swer any  rational  question  that  can  be  put  with  respect  to 
the  position.  Accordingly,  since  any  position  in  Space  is 
known  completely  when,  and  only  when,  three  independent 
data  are  known  about  it,  we  say  that  Space  is  triply  or  three- 
fold extended^  or  has  three  dimensions.  The  dimensions  are 
any  three  independent  things  that  it  is  necessary  and  suffi- 
cient to  know  about  any  position  in  Space,  as  of  the  pin- 
point or  of  Jupiter,  in  order  to  know  exactly  where  it  is. 

3.  But  with  respect  to  the  outside  of  the  earth,  viewed 
as  a  level  sheet  of  water,  we  have  seen  that  only  two  data, 
as  of  latitude  and  longitude,  are  necessary  and  sufficient 
to  fix  any  position  on  it ;  neither  are  more  than  two  inde- 
pendent data  possible  ;  all  other  knowledge  about  the  posi- 
tion is  involved  in  the  knowledge  of  these  two  data  about 
it.  Accordingly  we  say  of  such  outside  of  the  earth  that  it 
is  doubly  or  two-fold  extended y  is  bi-dimensional,  or  has  two 
dimensions ;  and  we  name  every  such  outside^  every  such 
bi-dimensional  region,  a  surface.  Such  is  the  top  of  the 
table  :  to  know  where  a  spot  is  on  it  we  need  know  two,  and 
only  two,  independent  facts  about  it,  as  how  far  it  is  from 
the  one  edge  and  how  far  from  the  other.  (Which  other? 
and  why?) 

We  see  at  once  that  a  surface  is  no  part  of  Space,  but  is 
only  a  border  (doubly  extended)  between  two  parts  of  Space. 
Thus,  the  whole  earth-surface  is  no  part  either  of  the  earth- 
space  or  of  the  air-space  around  the  earth,  but  is  the  boun- 
dary between  them.  A  soap-bubble  floating  in  the  air  is 
not  a  surface ;  though  exceedingly  thin,  it  has  some  thick- 
ness and  occupies  a  part  of  space ;  the  outside  of  the  film 


6  GEOMETRY. 

is  a  surface,  and  so  is  the  inside,  and  these  are  kept  apart 
by  the  film  itself.  If  the  film  had  no  thickness,  the  outside 
and  the  inside  would  fall  together,  and  the  film  would  be  a 
surface ;  namely,  the  outside  of  the  Space  within  and  the 
inside  of  the  Space  without. 

4.  Consider  now  once  more  this  earth- surface,  still  viewed 
as  a  smooth  level  sheet  of  water.  From  Geography  we 
learn  that  there  are  two  extreme  positions  on  this  surface 
that  are  called  poles  and  that  do  not  move  at  all  as  the 
earth  spins  round  on  her  axis.  We  also  learn  that  there  is 
a  certain  region  of  positions  just  midway  between  these 
poles  and  called  the  Equator.  This  Equator  is  no  part  of 
the  surface ;  it  is  only  a  border  or  boundary  between  two 
parts  of  the  globe-surface,  which  are  called  hemispheres. 
To  know  where  any  position  is  on  this  border,  it  is  neces- 
sary and  sufficient  to  know  o?ie  thing  about  it,  namely,  its 
longitude ;  neither  is  any  other  independent  knowledge 
about  the  position  possible  ;  all  other  knowledge  is  involved 
in  this  one  knowledge.  Accordingly  we  say  of  this  border, 
the  Equator,  that  it  is  simply  extended,  or  has  one  dimension 
only.  Every  such  one-dimensional  border  is  called  a  line, 
and  its  one  dimension  is  named  length.  A  line,  then,  has 
length,  but  neither  breadth  nor  thickness. 

5.  Lastly,  consider  a  part  of  a  line,  as  of  the  Equator, 
say  between  longitudes  40°  and  50°.  The  ends  of  this  part 
bound  it  off  from  the  rest  of  the  equator,  but  they  them- 
selves form  no  part  of  the  Equator.  They  are  called  points  ; 
they  have  position  merely,  but  no  extent  of  any  kind,  neither 
length  nor  breadth  nor  thickness,  —  they  are  wholly  non- 
dimensional. 


INTRODUCTION.  7 

*6.  It  is  noteworthy  that  extents,  or  regions,  are  bounded 
by  extents  of  fewer  dimensions,  and  themselves  bound 
extents  of  more  dimensions.  Thus,  lines  are  bounded  by 
points,  and  themselves  bound  surfaces ;  surfaces  are  bounded 
by  lines,  and  themselves  bound  spaces  ;  spaces  are  bounded 
by  surfaces,  and  themselves  bound  —  what?  If  anything 
at  all,  it  must  be  some  extent  of  still  higher  order,  oi  four 
dimensions.  But  here  it  is  that  our  intuition  fails  us ;  our 
vision  of  the  world  knows  nothing  of  any  fourth  dimension, 
but  is  confined  to  three  dimensions.  If  there  be  any  such 
fourth  dimension,  we  can  know  nothing  of  it  by  intuition ; 
we  cannot  imagine  it.  In  music,  however,  we  do  recognize 
four  dimensions  :  in  order  to  know  a  note  completely,  to 
distinguish  it  from  every  other  note,  we  must  know  four 
things  about  it :  its  pitch,  its  intensity,  its  length,  its  timbre, 
—  how  high  it  is,  how  loud  it  is,  how  long  it  is,  how  rich 
it  is.  While,  then,  extents  of  higher  dimensions  may  be 
unimaginable,  they  are  not  at  all  unreasonable. 

This  doctrine  of  dimensions  is  of  prime  importance,  but 
rather  subtile  ;  let  not  the  student  be  disheartened,  if  at  first 
he  fail  to  master  it. 

6.  We  may  see  and  handle  bodies,  which  occupy  portions 
of  Space ;  but  not  so  surfaces,  lines,  points,  which  occupy 
no  Space,  but  are  merely  regions  in  Space.  Here  we  must 
invoke  the  help  of  the  logical  process  called  abstraction, 
i.e.  withdrawing  attention  from  certain  matters,  disregard- 
ing them,  while  regarding  others.  A  sheet  of  paper  is  not 
a  surface,  but  a  body  occupying  Space.  However  thin,  it 
yet  has  some  thickness.  But  in  thinking  about  it  we  may 
leave  its  thickness  out  of  our  thoughts,  disregard  its  thick- 
ness altogether ;  so  it  becomes  for  our  thought,  though  not 
for  our  senses  or  imagination,  a  surface.    The  like  may  be 


8 


GEOMETRY. 


said  of  the  film  of  the  soap-bubble.  Again,  consider  the 
pointer.  It  is  a  body  or  solid,  not  only  long,  but  wide  and 
thick ;  it  occupies  Space.  It  is  neither  line  nor  surface. 
But  we  may,  and  do  often,  disregard  wholly  two  of  its 
dimensions,  and  attend  solely  to  the  fact  that  it  is  long. 
Thus  it  becomes  for  our  thought  a  line,  though  not  for  our 
senses  or  imagination.  So  the  mark  made  with  chalk  or 
ink  or  pencil  is  a  body,  triply  extended ;  but  we  disregard 
all  but  its  length,  and  it  becomes  for  our  reason  a  line. 
Lastly,  we  make  a  dot  with  pen  or  chalk  or  pencil ;  it  is  a 
body,  tri-dimensional,  occupying  Space.  But  we  may  dis- 
regard all  its  dimensions,  and  attend  solely  to  the  fact  that 
it  has  position,  that  it  is  here,  and  not  there.  So  it  becomes 
in  our  thought  a  point.  By  such  abstraction  the  earth,  the 
sun,  the  stars,  the  planets,  may  all  be  treated  as  points. 


Fig.  I. 


7.  Inasmuch  as  Space  is  continuous,  there  may  also  be 
continuous  surfaces  and  lines ;  and  the  only  surfaces  and 
lines  treated  in  this  book  are  continuous,  without  holes,  gaps, 
rents,  breaks,  or  interruptions  of  any  kind  in  their  extent.    , 

It  is  important  to  note  that  in  passing  from  any  position  A 


INTRODUCTION.  9 

to  another  B  on  a  continuous  line,  a  moving  point  P  must 
pass  through  a  complete  series  of  intermediate  positions ; 
/>.  there  is  no  position  on  the  line  between  A  and  B  that 
the  point  P  would  not  assume  in  going  from  A  to  B. 
(Fig.  I.) 

*8.  Starting  from  the  notion  of  Space,  we  have  attained 
the  notions  of  surface,  line,  and  point,  in  two  ways :  by  treat- 
ing them  as  borders,  and  by  the  process  of  abstraction.  But 
we  may  reverse  this  order  and  attain  the  notions  of  line, 
surface,  and  solid  or  space  from  the  notion  of  point,  with 
the  help  of  the  notion  of  motion,  thus :  Let  a  point  be 
defined  as  \izs\x\g  position  without  parts  or  inagnitude  of  any 
kind.  Let  it  move  continuously  through  Space  from  the 
position  A  to  the  position  B.  To  know  where  it  is  at  any 
stage  of  its  motion  along  any  definite  path,  it  is  necessary 


Fig.  2. 

and  sufficient  to  know  one  thing ;  namely,  how  far  it  is 
from  A.  Hence  its  path  is  a  6? w^- dimensional  extent,  or 
what  we  call  a  line. 


10  GEOMETRY. 

Now  let  a  line  move  in  any  definite  way  from  any  position 
Q  to  any  other  position  R.  To  know  the  position  of  any 
point  of  its  path,  it  is  necessary  and  sufficient  to  know  two 
things ;  namely,  the  position  of  the  point  on  the  moving  line 
and  the  position  of  the  moving  line  itself:  hence  the  path 
of  the  line  is  a  /z^/^-dimensional  extent,  which  we  have 
already  named  a  surface.     (Fig.  2.) 

Now  let  a  surface  move  in  any  definite  way  from  any 
position  U  to  any  other  position  V.  To  know  the  position 
of  any  point  on  its  path  it  is  necessary  and  sufficient  to 
know  three  things  about  it ;  namely,  its  position  on  the 
moving  surface  (which,  we  know,  counts  as  two  things)  and 
the  position  of  the  moving  surface  itself.  Hence  the  path 
of  the  surface  is  a  //jr^^f-dimensional  extent,  which  we  have 
already  named  a  solid  or  a  part  of  space. 

Now,  if  we  let  a  solid  move,  what  will  its  path  be? 
Naturally  we  should  expect  it  to  be  a  /^z^r- dimensional 
extent,  but  no  such  extent  is  yielded  in  our  experience  by 
any  motion  of  a  solid  —  the  path  of  a  solid  is  nothing  but 
a  solid.  The  explanation  of  the  apparent  inconsistency  is 
very  simple,  to-wit :  A  piece  of  a  line  traces  out  a  surface 
only  when  it  moves  out  from  the  line  itself,  —  if  one  part 
were  to  slip  round  on  another  part  of  the  same  line,  it  would 
trace  out  no  surface  at  all  as  its  path ;  likewise,  a  piece  of 
a  surface  traces  out  a  solid  as  its  path  only  by  moving  out 
from  the  surface  itself,  —  if  one  part  were  to  slip  round  on 
another  part  of  the  surface,  it  would  trace  out  no  solid  at 
all  as  its  path.  So,  if  a  piece  of  our  space  could  move  out 
from  space  itself,  it  would  trace  out  a  four-fold  extent  as  its 
path ;  in  fact,  however,  no  part  of  space  can  move  out  from 
space ;  on  the  contrary,  it  can  only  slip  along  in  space,  from 
one  part  of  space  to  another,  and  hence  does  not  trace  out 
any  four-fold  extended  path. 


INTROD  UCTION.  1 1 

9.  Space,  we  have  seen,  is  homceoidal,  everywhere  alike. 
We  naturally  inquire  :  Is  there  any  homueoidal  surface  ?  In 
general,  surfaces  are  certainly  not  homceoidal.  Consider  an 
egg-shell,  and  by  abstraction  treat  it  as  a  surface.  It  is  not 
alike  throughout ;  the  ends  are  not  like  each  other,  and 
neither  is  like  the  middle  region.  Suppose  a  piece  cut  out 
anywhere ;  if  slipped  about  over  the  rest  of  the  shell,  this 
piece  will  not  fit.  But  now  consider  a  smooth  round  ball 
covered  with  a  thin  rigid  film,  and  treat  this  film  as  a  sur- 
face, by  disregarding  its  thickness.  Suppose  a  piece  of  the 
film  cut  out  and  slipped  round  over  the  rest  of  the  film  :  the 
piece  will  fit  everywhere  perfectly,  the  surface  is  homceoidal ; 
it  is  called  a  sphere-surface. 

N.B.  The  precise  definition  of  this  surface  is  that  all 
its  points  are  equidistant  from  a  point  within,  called  the 
centre.  Suppose  a  rigid  bar  of  any  shape,  pointed  at  both 
ends,  and  movable  about  one  end  fixed  at  a  point ;  then 
the  other  end  will  move  always  on  a  sphere-surface,  which 
is  the  whole  region  where  the  moving  end  may  be.  Since 
Space  is  homceoidal  around  the  fixed  point,  the  surface 
everywhere  equidistant  from  the  point  is  also  homceoidal. 

Now  turn  over  the  piece  cut  out  of  this  spherical  film 
and  slip  it  about  the  film  :  it  no  longer  fits  anywhere  at  all 
—  the  surface  is  homceoidal,  but  not  reversible. 

10.  But  now  consider  a  fine  mirror  covered  with  a  deli- 
cate film,  which  by  abstraction  we  treat  as  a  surface.  Sup- 
pose a  piece  cut  out  of  the  film  and  slipped  about  over  it : 
the  piece  fits  everywhere ;  turn  it  over,  re-apply  it,  and  slip 
it  alK)Ut :  it  still  fits  everywhere  —  the  surface  is  both  homa- 
oidalzxA  reversible  ;  it  is  called  a  plane-surface. 

*  N.B.  A  precise  definition  of  this  surface  is  the  following  : 
Take  two  points  A  and  B  and  suppose  two  equal  spherical 


12  GEOMETRY. 

bubbles  formed  about  A  and  B  as  centres.     Let  them  ex- 
pand, always  equal  to  each  other,  until  they  meet,  and  still 
,  ^     keep  on  expanding.    The  line  where 

^  S   the  equal  (Fig.  3)  spherical  bubbles, 

^^'  ^"  regarded  as  surfaces,  meet,  has  all 

its  points  just  as  far  from  A  as  from  B.  As  the  bubbles 
still  expand,  this  line,  with  all  its  points  equidistant  from  A 
and  B,  itself  expands  and  traces  out  a  plane  as  its  path 
through  Space. 

Hence  we  may  define  \he  plane  as  the  region  (or  surface) 
where  a  point  may  be  that  is  equidistant  from  two  fixed 
points.  Instead  of  region  it  is  common  to  say  locus,  i.e. 
place.  Briefly,  then,  a  plane  is  the  locus  of  a  point  equidis- 
tant from  two  fixed  points.  It  is  evident  that  the  plane,  as 
thus  defined,  is  reversible ;  for  since  the  bubbles  about  A 
and  B  are  all  the  time  precisely  equal,  to  exchange  A  and 
B^  or  to  exchange  the  sides  of  the  plane,  will  make  no  dif- 
ference whatever.  Thus  the  plane  cuts  the  Space  evenly 
half  in  two ;  and  since  Space  itself  is  homoeoidal,  so  also  is 
this  section  or  surface  that  halves  it  exactly.  The  superiority 
of  this  definition  consists  in  its  not  only  telling  what  surface 
the  plane  is,  but  also  making  clear  that  there  actually  is  such 
a  surface. 

1 1 .  The  mirror  is  the  nearest  approach  that  we  can  make 
to  a  perfect  plane  surface ;  the  blackboard  is  not  plane,  it 
is  rough  and  warped ;  but  we  shall  disregard  all  its  uneven- 
ness  and  treat  it  as  a  plane  extended  through  Space  without 
end.  Any  surface  may  be  dealt  with  as  a  plane  by  abstrac- 
tion, being  thought  as  ho?nceoidal  and  reversible. 

12.  On  this  board,  regarded  as  a  plane,  we. draw  a  chalk- 
mark,  abstract  from  all  its  dimensions  but  its  length,  and 


INTRODUCTION. 


13 


treat  it  as  a  line.     This  line  is  plainly  not  alike  throughout ; 
a  piece  cut  out  and  slipped  along  it  will  not  fit  (Fig.  4). 


FUJ.  4. 

But  here  is  a  line  homceoidal,  alike  in  all  its  parts ;  it  is 
drawn  with  a  pair  of  compasses  and  is  called  a  circle 
(Fig.  5).  One  point  of  the  compasses  is  held  fast  at  the 
centre  Oy  while  the  other  traces  out  the  circle  as  its  path  in 


B 


Fig.  s. 


•  W 

Fig.  6. 


the  plane.  The  circle  is  the  locus  of  a  point  in  the  plane 
equidistant  from  a  fixed  point.  Since  the  plane  is  homce- 
oidal, so  too  is  this  circle  (see  Art.  10)  ;  a  piece,  called  an 
arc,  cut  out  and  slipped  round  will  everywhere  fit  on  the 
circle.  But  turn  it  over  and  slip  it  round,  —  it  fits  nowhere  ; 
the  circle  is  not  reversible.     It  divides  the  plane  into  two 


14  GEOMETRY, 

parts,  not  halves,  that  are  not  ahke  along  the  dividing  line. 
But  now  suppose  a  perfectly  flexible  string  fastened  at  S  and 
stretched  by  a  weight  W.  Its  length  only  being  regarded, 
it  is  a  line  homoeoidal,  alike  throughout,  and  also  reversible ; 
any  part  AB  will  not  only  fit  perfectly  anywhere  on  it,  but 
will  also  fit  when  reversed,  turned  end  for  end.  Such  a 
line  is  called  right,  or  straight,  or  direct,  or  a  ray.  Extended 
indefinitely,  it  cuts  the  whole  plane  into  two  halves  pre- 
cisely alike  along  the  ray  itself. 

*N.B.  The  common  line  where  the  two  spherical  bub- 
bles of  Art.  lo  meet  is  a  circle,  for  it  is  plainly  precisely 
alike  all  around ;  it  is  homoeoidal,  being  the  intersection  of 
two  homoeoidal  surfaces,  namely,  the  two  equal  sphere- 
surfaces  ;  it  is  also  in  a  plane,  and  in  fact  traces  out  the 
plane  by  its  expansion  as  the  bubbles  expand. 

To  get  accurately  the  notion  of  the  ray  or  straight  line, 
we  need  another  point  C,  and  a  third  expanding  bubble 
always  equal  to  those  about  A  and  B.  The  circular  inter- 
section of  the  bubbles  about  A  and  B  will  trace  out  one 
plane  ;  of  those  about  B  and  C  will  trace  out  another  plane ; 
of  those  about  C  and  A  will  trace  a  third  plane.  All  the 
points  where  the  first  two  planes  intersect  will  be  equidistant 
from  A  and  B  and  C,  and  no  other  points  will  be;  the 
same  may  be  said  of  all  points  where  the  second  and  third 
planes  meet,  and  of  all  points  where  the  third  and  first  meet ; 
hence  all  three  of  the  planes  meet  together,  and  they  meet 
only  together.  Also,  the  line  where  they  meet  has  every 
one  of  its  points  equidistant  from  all  the  thre5  points,  A,  B, 
C ;  hence  it  is  the  locus  0/  a  point  equidistant  from  three 
fixed  points.  Moreover,  it  is  homoeoidal  and  reversible, 
since  it  is  the  intersection  of  two  planes,  which  are  homoe- 
oidal and  reversible ;  hence  it  is  what  we  call  a  straight 
line,  or  right  line,  or  ray. 


INTRODUCTION.  IS 

13.  We  may  now  define  : 

A  sphere- surface  is  the  locus  of  a  point  at  a  fixed 
distance  from  a  fixed  point.  It  is  homceoidal,  but  not 
reversible. 

A  plane  is  the  locus  of  a  point  equidistant  from  two  fixed 
points.     It  is  both  homceoidal  and  reversible. 

A  ray  is  the  locus  of  a  point  equidistant  from  three  fixed 
points.  It  is  both  homceoidal  and  reversible  ;  it  is  also  the 
intersection  of  two  planes. 

A  circle  is  the  locus  (or  path)  of  a  point  in  a  plane  at  a 
fixed  distance  from  a  fixed  point.  It  is  homceoidal,  but  not 
reversible.  It  is  also  the  locus  (or  path)  of  a  point  in  space 
at  a  fixed  distance  from  two  points ;  it  is  also  the  intersec- 
tion of  two  equal  sphere-surfaces.* 

14.  It  is  only  with  the  foregoing  figures  and  combinations 
of  them  that  we  have  to  deal  in  this  book.  Circles  and  rays 
may  be  drawn  with  exceeding  accuracy,  but  any  lines,  how- 
ever roughly  drawn,  may  answer  our  logical  purposes  as 
well  as  the  most  accurately  drawn ;  we  have  only,  by 
abstraction,  to  treat  them  as  having  the  character  of  the 
lines  in  question. 

Circles  and  sphere-surfaces  are  unbounded,  without  be- 
ginning or  end,  but  both  are  finite :  we  shall  learn  how  to 
measure  them. 


*  In  the  foregoing  free  use  has  been  made  of  the  notion  of  equidistance  without 
formal  definition,  because  of  its  familiarity.  We  may,  however,  say  precisely:  \{  A 
and  B  be  two  points,  the  ends  of  a  rigid  bar  of  any  shape,  and  if  /I  be  held  fast,  then 
all  the  points  on  which  B  can  fall  are  equidistant  from  A ,  and  no  other  points  arc 
equidistant  with  them.  They  all  lie  on  a  closed  surface,  called  a  sphere-surface. 
All  points  within  this  surface  are  said  to  be  less  distant,  and  all  points  without  are 
said  to  be  more  distant,  from  A  than  B  is.  Herewith,  then,  we  tell  exactly  what 
we  mean  by  equidistant,  less  distant,  and  more  distant,  but  we  make  no  attempt  to 
define  distatue  in  general,  which  is  difficult  and  unnecessary  to  our  purpose. 


16  GEOMETRY, 

15.  Any  geometric  element  or  combination  of  geometric 
elements,  as  points,  lines,  surfaces,  is  called  a  geometric 
figure.  It  is  a  fundamental  assumption,  justified  by  experi- 
ence, that  space  is  homoeoidal,  that  figures  or  bodies  are  not 
affected  in  size  or  shape  by  change  of  place.  Two  figures 
that  may  be  fitted  exactly  on  each  other,  or  may  be  thought 
so  fitted,  are  called  congruent.  Any  two  points,  lines,  or 
parts  of  the  two  figures,  that  fall  upon  each  other  in  this 
superposition  are  said  to  correspond.  It  is  manifest  that  all 
planes  are  congruent  and  all  rays  are  congruent.  Rays  and 
planes  are  unbounded,  but  whether  or  not  they  are  finite  is 
a  question  that  we  are  unable  to  answer. 

16.  Any  part  of  a  circle  or  ray,  as  AB,  is  bounded  by 
two  end-points,  A  and  B,  and  is  finite ;  the  one  is  named 
an  arc  (Fig.  5),  the  other  a  tract,  sect,  or  line-segment. 
Each  is  denoted  by  the  two  letters  denoting  the  ends,  as 
the  tract  AB,  the  arc  AB.  Sometimes  it  is  important  to 
distinguish  these  end-points  as  beginning  and  end  proper ; 
we  do  this  by  writing  the  letter  at  the  beginning  first. 


A' 

B' 

A 

B 

C 

D 
D' 

A' 

B' 

I-T- 

E 
Fig.  7. 

17.  Two  tracts,  AB  and  AB\  are  called  equal  when  the 
end-points  of  the  one  may  be  (Fig.  7)  simultaneously  fitted 
on  the  end-points  of  the  other. 

If  we  have  a  number  of  tracts,  AB,  CD,  EF,  etc.,  and 
we  lay  off  successively  on  a  ray  tracts  A^B\  C D\  E^P,  etc., 


INTKODUCriON.  17 

respectively  equal  to  AB,  CDy  EF,  etc.,  the  end  of  the  first 
being  the  beginning  of  the  second,  and  so  on,  while  no  part 
of  one  falls  on  any  part  of  another,  we  are  said  to  add  or 
sum  the  tracts  ABj  etc.  Each  is  called  an  addend  or  sum- 
mand,  and  the  whole  tract  from  first  beginning  to  last  end 
is  called  the  sum. 

Equality  is  denoted  by  the  bars  ( = )  between  the  equals, 
as  AB  =  CD. 

i8.  If,  when  the  beginning  A  is  placed  on  the  beginning 
C,  the  end  B  does  not  fall  on  the  end  Z>,  the  tracts  are 
unequal,  and  we  write  AB  ^  CD.  If  B  falls  between  C 
and  A  then  AB  is  called  less  than  CD,  AB<CD\  but  if 
D  falls  between  A  and  B,  then  AB  is  called  greater  than 
CD,  AB  >  CD.  In  either  case,  the  tract  BD  or  DB, 
between  the  two  ends  of  the  tracts,  whose  beginnings  coin- 
cide, is  called  the  difference  of  the  two  tracts,  and  we  are 
said  to  subtract  the  one  from  the  other.  Ordinarily  we 
mention  the  greater  tract  first  in  speaking  of  difference. 

19.  The  symbols  of  addition  and  subtraction  are  +  and 
—  (plus  and  minus),  thus  : 

AB-^-  CD^AD  and  AB  -  CD=BD. 

It  is  important  to  note  here  the  order  of  the  letters.  In 
summing  a  number  of  tracts,  as  AB,  CD,  EF,  etc.,  to  A'Z, 


Fig.  8. 

we  have  AB  +  CD -\- EF--  +  KL  =  AL  (Fig.  8).  The 
order  of  the  summands  is  indifferent,  and  this  important 
fact  is  called  the  Commutative  Law  of  Addition.     Thus 

AB+  CD+EF=AB^-EF-\-  CD=EF-\-AB-\-  CD,  etc. 


18  GEOMETRY. 

'  20.  When  beginning  and  end  of  a  tract  or  of  any  mag- 
nitude are  exchanged,  the  tract  or  magnitude  is  said  to  be 
reversed,  and  the  reverse  is  denoted  by  the  sign  — .  Thus 
the  reverse  of  AB  is  BA,  or  AB  =  —  BA.  If  we  add  a 
magnitude  and  its  reverse,  the  sum  is  o,  or 

AB  -h{-AB)  =  AB-\-BA  =  o. 

The  same  result  o  is  obtained  by  subtracting,  from  a  magni- 
tude, itself  or  an  equal  magnitude ;  and,  in  general,  it  is 
plain  that  to  subtract  CB>  yields  the  same  result  as  to  add 
(Fig.  9)  the  reverse  B>C.     The  reverse  of  a  magnitude  is 


A  B 

C D C 

H 1 J^ 

A  B 


Fig.  9. 

often  called  its  negative,  the  magnitude  itself  being  called 
its  positive. 

Similar  rules  hold  for  adding  and  subtracting  arcs  of  a 
circle  or  of  equal  circles. 

ANGLES. 

21.  The  indefinite  extent  of  a  ray  on  one  side  of  a  point 
O,  as  OA,  is  called  a  half-ray :  it  has  a  beginning  O,  but  no 
end.  Two  half-rays,  OA  and  OA\  which  together  make  up 
a  whole  ray,  are  called  opposite  or  counter  (Fig.  10). 

Now  let  two  half-rays,  OA  and  OB,  have  the  same  be- 
ginning O ;  the  opening  or  spread  between  them  is  a  mag- 
nitude :  it  may  be  greater  or  less.  Suppose  OA  and  OB  to 
be  two  very  fine  needles  pivoted  at  O ;  then  OB  may  fall 
exactly  on  OA,  or  it  may  be  turned  round  from  OA ;  and 


JXTKODUCriON.  19 

the  amount  of  turning  from  OA  to  OB,  or  the  spread 
between  the  half- rays,  is  called  the  angle  between  them. 
We  may  denote  it  by  a  Greek  letter,  as  «,  ^vritten  in  it ;  or 
by  a  large  Roman  letter,  as  O,  at  its  vertex  (where  the  half- 
rays  meet)  ;  or  by  three  such  letters,  as  AOB,  the  middle 


Fig.  io. 

one  being  at  the  vertex,  the  other  two  anywhere  on  the  half- 
rays.    The  symbol  for  angle  is  ^. 

22.  The  angle  is  perfectly  definite  in  size,  it  has  two 
ends  or  boundaries ;  namely,  the  two  half-rays,  sometimes 
called  arms.  When  we  would  distinguish  these  arms  as 
beginning  and  end,  we  mention  the  letter  on  the  beginning- 
arm  first,  and  the  letter  on  the  end-arm  last;  thus,  A  OB; 
here  OA  is  the  beginning  and  OB  the  end  of  the  angle. 

Exchanging  beginning  and  end  reverses  the  angle ;  thus, 
BOA  =  -AOB. 

23.  Two  angles  whose  ends  or  arms  may  be  made  to  fit 
on  each  other  simultaneously  are  named  equal ;  they  are  also 
congruent.  Two  angles  whose  arms  will  not  fit  on  each  other 
simultaneously  are  unequal;  and  that  is  the  less  angle  whose 
end-arm  falls  within  the  other  angle  when  their  beginnings 


20 


GEOMETRY. 


coincide;    the    other   is  the  greater;    thus,    AOB>  AOC 
(Fig.  II). 

24.  We  sum  angles  precisely  as  we  sum  tracts ;  we  lay 
off  a,  13,  etc.,  around  O,  making  the  end  of  each  the  begin- 
ning of  the  next :  the  angle  from  first  beginning  to  last  end 


Fig.  II. 

is  the  stmt.  So,  too,  in  order  to  subtract  /?  from  «,  lay  off 
/8  from  the  beginning  towards  the  end  of  a ;  the  angle  from 
the  end  of  (3  to  the  end  of  a  is  the  difference,  a  —  fi.  Or 
we  may  add  to  a  the  reverse  (or  negative)  of  ^ :  the  sum 
will  be  «  +  (-/5)  or  a-jS  (Fig.  12).^ 

^  It  is  important  to  note  the  close  correspondence  of  tract  and  angle:  the  former  is 
related  to  points  as  the  latter  is  to  rays  (or  half-rays).  The  tract  is  the  simplest 
magnitude  that  lies  between  points,  that  distinguishes  them  and  keeps  them  apart; 
likewise  the  angle  is  the  simplest  magnitude  that  lies  between  rays  (in  a  plane),  that 
distinguishes  them  and  keeps  them  apart.  So,  too,  we  define  equality  and  inequality 
among  tracts  and  among  angles,  quite  similarly,  and  without  being  compelled  before- 
hand to  form  the  notion  of  the  size  either  of  a  tract  or  of  an  angle.  We  may  now 
define  the  distance  between  two  points  to  be  the  tract  between  them,  and  the  dts- 
tatice  between  two  (half-) rays  to  be  the  angle  between  them,  leaving  for  future 
decision  which  tract  and  which  angle  if  there  should  prove  to  be  several. 


INTRODUCTION. 
'D 


21 


Fig.  12. 


AXIOMS. 

25.  At  this  stage  we  must  recognize  and  use  certain  dic- 
tates or  irresoluble  facts  of  experience,  called  axioms. 
{'kiniiiuimQ^ns  somethifig  worthy,  like  the  Latin  dignitas ;  in 
fact,  older  writers  use  dignity  in  the  sense  of  axiom.  But 
Euclid's  phrase  is  kqlvqx  twouu  =  common  notions.^  Some 
have  no  special  reference  to  Geometry,  but  pervade  all  of 
our  thinking  about  magnitudes ;  such  are 

(i)  Things  equal  to  the  same  thing  are  equal  to  each 
other. 

(2)  If  equals  be  added  to,  subtracted  from,  multiplied  by, 
or  divided  by,  ccjuals,  the  results  will  be  equal. 


22  GEOMETRY. 

(3)  If  equals  be  added  to  or  subtracted  from  unequals, 
the  latter  will  remain  unequal  as  before. 

(4)  The  whole  equals,  or  is  the  sum  of,  all  its  distinct 
parts,  and  is  greater  than  any  of  its  parts. 

(5)  If  a  necessary  consequence  of  any  supposition  is 
false,  the  supposition  itself  is  false. 

Others  concern  Geometry  especially,  as  : 

(6)  All  planes  are  congruent. 

(7)  Two  rays  can  meet  in  only  one  point. 

The  extremely  important  axiom  (7)  may  be  stated  in 
other  equivalent  ways,  thus  :  Two  rays  cannot  meet  in  two 
or  more  points ;  or,  Two  rays  cannot  have  two  or  more 
points  in  common ;  or.  Only  one  ray  can  go  through  two 
fixed  points ;  or,  A  ray  is  fixed  by  two  points. 

26.  A  statement  or  declaration  in  words  is  called  -a,  p7'op- 
osition.  The  propositions  with  which  we  have  to  deal  state 
geometric  facts  and  are  also  called  Theorems  (Oeiopyjfia,  from 
OewpeLv,  to  look  at,  means  the  product  of  metital  contempla- 
tioTi).  Propositions  are  often  incorrect;  theorems,  never. 
Subordinate  facts,  special  cases  of  general  facts,  and  facts 
immediately  evidenced  from  some  preceding  facts,  are 
called  Corollaries  or  Porisms  {■Tropi(Tixa  =  dedi/ctio?i). 

We  may  now  proceed  to  investigate  lines  and  angles,  and 
find  out  what  we  can  about  them.  The  first  and  simplest 
things  we  can  learn  concern 


Tn.  I.]  CONGRUENCE,  23 


CONGRUENCE. 

27.    Theorem  I. — All  rays  are  congruent. 

Proof.  Let  L  and  Z'  be  any  two  rays  (Fig.  13).  On 
L  take  any  two  points,  A  and  B ;  on  L'  take  any  two 
points,  A'  and  B',  so  that  the  tract  AB  shall  equal  the 
tract  A'B'.  Think  of  Z  and  Z'  as  extremely  fine  rigid 
spider-threads,  and  in  thought  place  the  ends  of  the  tract 
AB  on  the  ends  of  the  tract  A'B',  A  on  A\  and  B  on  B'. 


B 


A  B 

Fig.  13. 

Then  A  and  A'  become  one  and  the  same  point,  and  B  and 
B'  become  one  and  the  same  point ;  through  these  two 
points  only  one  ray  can  pass  (by  Axiom  7)  :  hence  Z  and 
Z',  which  go  through  these  two  points,  now  become  one 
and  the  same  ray ;  that  is,  they  fit  precisely,  they  are  con- 
gruent. Quod  erat  demonstrandum  =  which  was  to  be 
/>rotred  =  oTTtp  iStL  Sci^ai,  —  the  solemn  Greek  formula; 
whereas  the  Hindu,  appealing  directly  to  intuition,  merely 
said  Pafya  —  Behold  ! 

28.  In  the  foregoing  proof  we  assumed  that  on  any  ray 
we  could  lay  off  a  tract  equal  to  a  given  tract,  or  that  on 
any  ray  we  could  find  two  points,  A  and  B,  as  far  apart  as 
two  other  points.  A'  and  B'.  This  assumption  that  something 
can  be  done,  is  called  a  Postulate  (alr-qfia),  I.e.  a  demand, 
which  must  be  granted  before  we  can  proceed  further. 
Actually  to  carry  out  the  construction,  we  need  a  pair  of 
compasses. 


24  GEOMETRY.  [Th.  11. 

29.  Theorem  II.  — If  two  points  of  a  ray  lie  in  a  certain 
plane,  all  points  of  the  ray  lie  in  that  plane. 

Proof.  Regard  the  surface  of  paper  or  of  the  blackboard 
as  a  plane,  and  suppose  it  covered  with  a  fine  rigid  film, 
itself  a  plane.  Let  L  be  any  ray  having  two  points,  A  and 
B,  in  this  plane.  Through  these  two  points  suppose  a 
second  plane  drawn  or  passed;  by  definition  (Art.  13)  it 
will  intersect  our  first  plane,  or  film,  along  a  ray  /;  this  ray 
/  goes  through  the  two  points,  A  and  B,  and  lies  wholly 
(with  all  its  points)  in  the  first  plane  ;  also  the  ray  L  goes 
through  A  and  B,  and  only  one  ray  can  go  through  the  same 
two  points,  A  and  B,  by  Axiom  7  ;  hence  L  and  /  are 
the  same  ray ;  but  /  has  all  its  points  in  the  first  plane ; 
hence  L  has  all  its  points  in  the  first  plane,     q.  e.  d. 

Query  :    What  postulate  is  assumed  in  this  proof  ? 

Corollary.  If  a  ray  turn  about  a  fixed  point  P,  and  glide 
along  a  fixed  ray  Z,  it  will  trace  out  a  plane  (Fig.  14). 


Fig.  14. 


For  it  will  always  have  two  points  —  namely,  the  fixed 
point  and  a  point  on  the  fixed  ray  —  in  the  plane  drawn 
through  the  fixed  point  and  the  fixed  ray. 


Th.  II.]  CONGRUENCE.  25 

Query:  What  postulate  is  here  implied? — Henceforth  it 
is  understood  that  all  our  points,  lines,  etc.,  are  complanar, 
/>.  lie  in  one  and  the  same  plane. 

30.  In  the  foregoing  Theorem  and  Corollary  we  observe 
clauses  introduced  by  the  word  if.  Such  a  clause  is  called 
an  Hypothesis,  i.e.  a  supposition.  The  result  reached  by 
reasoning  from  the  hypothesis  and  stated  immediately  after 
the  hypothesis,  is  called  the  Conclusion. 

31.  All  logical  processes  consist  in  one  or  both  of  two 
things  :  the  formation  of  concepts,  as  of  lines,  surfaces, 
angles,  etc.,  and  the  combination  of  these  concepts  into 
propositions.  Geometric  concepts  are  remarkable  for  their 
perfect  clearness  and  precision  —  we  know  exactly  what  we 
mean  by  them  ;  this  cannot  be  said  of  many  other  concepts, 
about  which  diverse  opinions  prevail,  as  in  Political  Economy. 
Hence  it  is  that  Geometry  offers  an  unequalled  gymnasium 
for  the  reason  or  logical  faculty.  We  shall  now  generate 
some  new  concepts.  Let  the  student  note  their  definiteness 
as  well  as  the  mode  of  their  formation. 

32.  Let  OA  and  OB  be  any  two  co-initial  half- rays, 
forming  the  angle  A  OB.  Think  of  OA  as  held  fast  and  of 
OB  as  turning  about  the  pivot  (9,  starting  from  the  position 
OA.     As  it  turns  (counter-clockwise),  the  (Fig.  15)  angle 


Fig.  15. 

AOB  increases.     Finally,  let  it  return  to  its  original  posi- 
tion, OA  ;  then  the  whole  amount  of  turning  from  the  upper 


26  GEOMETRY.  [Th.  III. 

side  of  OA  back  to  the  under  side  of  OA,  or  the  full  spread 
around  the  point  O,  is  called  a  /////  angle  (or  round  angle, 
or  c'lrcum-dcci^t,  or  perigon) ,  Think  of  a  fan  opened  until 
the  first  rib  falls  on  the  last.  —  Note  that  the  upper  and 
under  sides  of  OA  are  exactly  the  same  in  position,  and  are 
distinguished  only  in  thought.  (Think  of  a  circular  piece 
of  paper  slit  straight  through  from  the  edge  to  the  centre.) 
The  like  may  be  said  of  the  two  sides  of  any  line  or  surface. 
We  can  now  prove 

33.    Theorem  III.  —  All  round  angles  are  congruent. 

Proof.  Let  AOB  and  AO'B'  (Fig.  16)  be  any  two 
round  angles.  Slip  the  half-ray  OA  down,  and  turn  it  till 
OA  falls  on  O'A^  \     they  will  fit  perfectly  (why?);     the 


Fig.  16. 

whole  round  angle  about  O  will  fit  perfectly  on  the  whole 
round  angle  about  O'  (why?);  hence  the  two  full  angles 
are  congruent,     q.  e.  d. 

N.B.  In  this  slipping  of  figures  about  in  the  plane,  it  is 
well  to  imagine  the  plane  to  consist  of  two  very  thin,  per- 
fectly rigid,  smooth  and  transparent  films ;  also,  to  imagine 
one  figure  drawn  in  the  lower  film  and  one  in  the  upper ; 
and  to  imagine  the  upper  slipped  about  at  will  over  the  lower. 

Query :  On  what  cardinal  property  of  the  plane  do  these 
considerations  hinge  ? 


Th.  IV.] 


CONGRUENCE. 


27 


34.    From  O  draw  any  half-ray  OA  ;   then  any  second 

half- ray  from  O,  as  OB,  will  (Fig.  17)  cut  the  round  angle 

AOA  into  two  angles,  A  OB  and  BOA.     The  end  OB  of 

the  first  falls  on  the  beginning,  OB,  of  the  second ;  while 

A 

B 


Fig.  17. 

the  end,  OA,  of  the  second  falls  on  the  beginning,  OA,  of 
the  first.  Hence  the  round  angle  AOA  is  their  sum,  by 
Art.  24. 

If  we  draw  any  number  of  half- rays,  OB,  OC,  etc.,  -"OL, 
the  round  angle  will  still  be  the  sum  of  the  consecutive 
angles  AOB,  BOC,  etc., --^LO A  ;  hence  we  discover  and 
enounce  this 

Theorem  IV.  —  T/ie  sum  of  the  consecutive  angles  about  a 
point  in  a  plane  is  a  round  angle. 

N.B.  We  cannot  apply  Axiom  i  immediately,  because 
we  do  not  know,  except  by  Art.  24,  what  is  meant  by  a  sum 
of  angles. 

35.  In  the  foregoing  article  we  have  exemplified  the 
erotetic,  questioning,  investigative  method,  in  which  the  result 


28  GEOMETRY.  [Th.  IV. 

is  not  announced  until  it  is  actually  discovered  and  estab- 
lished. In  Theorems  I.,  II.,  III.,  on  the  other  hand,  the 
dogmatic  procedure  was  illustrated,  the  fact  or  proposition 
being  announced  beforehand,  while  the  demonstration  fol- 
lowed after.  Each  method  has  its  merits,  and  we  shall 
employ  both. 

36.  As  OB  turns  round  from  the  upper  to  the  under 
side  of  OA,  the  angle  AOB  begins  by  being  less  than  BOA 
and  ends  by  (Fig.  19)  being  greater  than  BOA.     The  plane 


Fig.  19. 

is  continuous,  the  turning  is  continuous,  the  change  m  size 
is  continuous ;  hence,  in  passing  from  the  stage  of  being 
less  to  the  stage  of  being  greater,  the  angle  has  passed 
through  the  intermediate  stage  of  being  equal;  let  OA^  be 
the  position  of  the  rotating  half-ray  at  this  stage  of  equality, 
then  AOA^  =  A'OA.  Two  equal  parts  making  up  a  whole 
are  called  halves;  hence  AOA'  and  A'OA  are  halves  of 
the  full  angle  AOA  ;  they  are  named  straight  (or  flat) 
angles. 

37.    Now,  —  Halves  of  equals  are  equal ; 

All    straight    angles    are    halves    of    equals 
(namely,  equal  round  angles)  ; 


Th.  VII.J  CONGRUENCE.  29 

Hence 

Theorem  V.  — All  straight  angles  are  equal. 

This  argument  here  given  /*//  extenso  is  a  specimen  of  a 
syllogism  ((rvAAoyto-/tAo«  =  computation  =  thinking  together). 
The  first  two  propositions  are  called  premisses,  the  third 
and  last,  in  which  the  other  two  are  thought  together,  is 
called  conclusion.  All  reasoning  may  be  syllogized,  but 
this  is  rarely  done,  as  being  too  formal  and  tedious. 

38.  Theorem  VI. — Two  counter  half- rays  bound  a 
straight  angle. 


O 

Fig.  20. 

For,  let  OA  and  OA'  be  two  such  counter  half- rays  (Fig. 
20)  forming  the  whole  ray  AA\  Turn  the  upper  half  of 
the  plane  film  round  O  as  pivot  until  the  upper  OA'  falls  on 
the  lower  OA-,  then,  since  the  ray  is  reversible,  the  ray 
A  A'  will  fit' exactly  on  the  ray  A^A;  i.e.  the  two  angles 
AOA'  and  A' OA  are  congruent  and  equal;  and  the  two 
compose  the  round  angle  A  OA ;  hence  each  is  half  of 
A  OA  ;  i.e.  each  is  a  straight  angle,     q.  e.  d. 

39.  Theorem  VII. — Conversely,  The  half-rays  bounding 
a  straight  angle  are  counter. 


P 

Fic;.  21. 


Let  OA  and  OA'  bound  a  straight  angle  (Fig.  21)  AOA' ; 
also  let  TB  and  J*B'  be  two  counter  half-rays ;    then  they 


30  GEOMETRY.  [Th.  VII. 

bound  a  straight  angle  BPB\  by  Theorem  VI.  Since  all 
straight  angles  are  congruent,  we  may  fit  these  two  on  each 
other ;  i.e.  we  may  fit  OA  and  OA^  on  PB  and  PB^ ;  but 
B B^  is  a  ray ;  so  then  is  AA^ ;  i.e.  OA  and  OA^  are 
counter,     q.  e.  d. 

40.  We  may  define  a  straight  angle  as  an  angle  bounded  by 
counter  half-rays.     Then  we  may  prove  Theorem  V.  thus  : 

The  ends  of  all  straight  angles  are  pairs  of  counter  half- 
rays  (or  form  whole  rays)  ; 

But  all  such  pairs  (or  whole  rays)  are  congruent  (by 
Theorem  I.) ; 

Therefore,  all  ends  of  straight  angles  are  simultaneously 
congruent. 

But  when  the  ends  of  angles  are  (simultaneously)  congru- 
ent, so  are  the  angles  themselves. 

Hence  all  straight  angles  are  congruent,     q.  e.  d. 

Here  the  first  conclusion,  introduced  by  "  therefore,"  is 
deduced  from  two  premisses  ;  but  the  second,  introduced  by 
"hence,"  is  apparently  deduced  from  only  one.  Only 
apparently,  however ;  for  one  premiss  was  understood  but 
not  expressed  ;  namely,  all  straight  angles  are  angles  whose 
ends  are  congruent.  Without  some  such  implied  additional 
premiss,  it  would  be  impossible  to  draw  the  conclusion. 
Such  a  maimed  syllogism,  with  only  one  expressed  premiss, 
is  called  an  enthymeme.  The  great  body  of  our  reasoning 
is  enthymematic.  We  shall  frequently  call  for  the  suppressed 
premiss  or  reason  by  a  parenthetic  question  (Why?). 

41.  Now  draw  two  rays,  LV  and  MM\  meeting  at  O. 
Each  divides  the  round  angle  about  O  into  two  equal 
straight  angles,  and  together  they  (Fig.  22)  form  four  angles 
«,  ^,  «',  y8'.  Two  angles,  as  «  and  /8,  that  have  a  common 
arm,  are  called  adjacent.     Accordingly  we  see  at  once  : 


Th.  IX.]  CONGRUENCE.  31 

Theorem  VIII.  —  Where  tiuo  rays  iniersect^  the  sum  of  two 

difjaccnt  angles  is  a  straight  angle. 


M 


Fig.  22. 

* 

Two  angles  whose  sum  is  a  straight  angle  are  called 
supplemental ;  two  angles  whose  sum  is  a  round  angle  we 
may  call  explemental.  Two  angles  as  «  and  «',  the  arms  of 
the  one  being  counter  to  the  arms  of  the  other,  are  called 
opposite^  or  vertical^  or  counter. 

Theorem  IX.  —  When  two  rays  meety  the  opposite  angles 
formed  are  equal. 

For  a-{-P  =  S  (a  straight  angle)  (why?)  ;  and  «'-f  ^  =  5 
(why?). 

Hence  u  +  p  =  u'  -{-  /3  (why?)  ;  therefore  a  =  a'.  Simi- 
larly let  the  student  show  that  ft  =  )S'.     q.  e.  d. 

An  important  special  case  is  when  the  adjacentSy  a  and  y8, 
are  equal.  Each  then  is  half  of  a  straight  angle,  and  there- 
fore one  fourth  of  a  round  angle  ;  and  each  is  called  a  right 
angle.  Now  let  the  student  show  that  if  «  =  ^,  then  «'  =  ^ 
and  a  —  /?',  or 

Corollary,  When  two  intersecting  rays  make  two  equal 
adjacent  angles y  they  make  all  four  of  the  angles  equal  (Fig. 
23). 

Def  Rays  that  make  right  angles  with  one  another  are 
called  normal  (or  perpendicular)  to  each  other.  N.B. 
The  normal  relation  is  mutual.     How? 


32 


GEOMETRY. 


[Th.  IX. 


Def.   Two  angles  whose  sum  is  a  right  angle  are  called 
complemental. 


Fig.  23. 

42.  Are  we  sure  that  through  any  point  on  a  ray  we  can 
draw  a  normal  to  the  ray?  Let  O  be  any  point  on  the 
ray  LV   (Fig  24).     Let  any  half-ray,  pivoted  at    O,  start 


R 

f 

•^ o^ 


Fig.  24. 


from  the  position  01.  and  turn  counter-clockwise  into  the 
position  0L\  At  first  the  angle  on  the  right  is  less  than  the 
angle  on  the  left,  at  last  it  is  greate?' ;  the  plane,  the  turning, 
and  the  angle  are  all  continuous ;  hence  in  passing  from  the 
stage  of  being  less  to  the  stage  of  being  greater,  it  passes 


Til.  X.]  CONGRUENCE.  33 

through  the  stage  of  equality.  Let  OR  be  its  position  in 
this  stage  ;  then  ^LOR^^  ROV  ;  i.e.  OR  is  normal  to 
LV.  Moreover,  in  no  other  position,  as  OSy  is  the  ray 
normal  to  LV  \  for  LOS  is  not  =  LOR  unless  OS  falls  on 
ORy  but  is  less  than  L  OR  when  OS  falls  within  the  angle 
LOR,  while  SOV  is  greater  than  LOR ;  hence  LOS  and 
SOV  are  not  equal ;  />.  6?^  is  not  normal  to  LL  when  (7.S 
falls  not  on  OR.  Similarly,  when  Z (96"  is  greater  than  LOR. 
Hence 

Theorem  X.  — Through  a  point  on  a  ray  one^  and  only  one^ 
ray  can  be  drawn  normal  to  the  ray, 

43.  Def.  A  ray  through  the  vertex  of  an  angle,  and 
forming  equal  angles  with  the  arms  of  the  angle,  is  called 
the  inner  Bisector  or  mid-ray  of  the  angle.  The  inner 
bisector  of  an  adjacent  supplemental  angle  is  called  the  outer 
bisector  of  the  angle  itself.  Thus  OI  bisects  innerly  and 
OE  bisects  outerly  the  angle  AOB  (Fig.  25). 


Exercise.    Prove  that  there  is  one  and  only  one  such  inner 
mid-ray. 


34 


GEOMETRY. 


[Th.  XI. 


44.  Theorem  XI.  —  The  inner  Bisector  of  an  angle 
bisects  also  its  cxp lenient  innerly. 

Proof.  Let  6>/ bisect  ^  AOB  innerly ;  then  '^AOI= 
^  lOB;  call  each  a;  then  a^-BOI'=:a  +  AOr  (why?)  ; 
take  away  «;  then  BOV^AOV  (why?)  ;  i.e.  the  ray  //' 
bisects  innerly  the  angle  BOA,  the  explement  of  A  OB. 
Show  that  the  angles  marked  «'  are  equal. 

45.  Theorem  XII.  —  The  inner  and  outer  Bisectors  of 
an  angle  are  tiormal  to  each  other. 

Proof.  Let  01  and  OE  bisect  (Fig.  25)  innerly  and 
outerly  the  angle  AOB.  Then,  by  definition,  the  angles 
marked  a  are  equal,  and  the  angles  marked  /?  are  equal ; 
also  the  sum  of  +  a  +  «  +  /8  +  y8  =  *S ;  hence  a  +  y8  =  J  6* ; 
or,  lOE  =  a  right  angle,     q.  e.  d. 

TRIANGLES. 


46.    Thus  far  we  have  treated  only  of  rays  intersecting  in 
a  single  point.     But,  in  general,  three  rays  Z,  J/,  N  (Fig. 


Fig.  26. 


26)  will  meet  in  three  points,  since  each  pair  will  meet  in 
one  point,  and  there  are  three  pairs  :  {MN) ,  (iVZ) ,  {LM) . 


Th.  XIII.] 


TRIANGLES. 


35 


Denote  these  points  by  A^  B^  C.  Then  the  figure  formed 
by  these  three  rays  is  called  a  triangle,  trigon,  or  three-side. 
Af  By  C  are  its  vertices ;  <f,  )8,  y  its  inner  angles ;  BC,  CA, 
AB,  its  inn^r  sides,  or  simply  its  sides.  Its  angles  and  sides 
are  called  its  parts.  It  is  the  simplest  closed  rectilinear 
figure,  and  most  important.  If  instead  of  taking  three 
rays  we  take  three  points  A^  B,  C,  then  we  may  join  them 
in  pairs  by  rays;  and  since  there  are  three  pairs,  BC,  CA, 
AB,  then  there  are  three  rays,  which  we  may  name  Z,  M,  N. 
Thus  we  see  that  three  points  determine  three  rays,  just  as 
three  rays  determine  three  points.  This  equivalent  deter- 
mination of  the  figure  by  the  same  number  of  points  as  of 
rays  makes  the  figure  unique  and  especially  important.  We 
denote  it  by  the  symbol  A.  We  now  ask,  When  are  two 
triangles  congruent? 

47.  Theorem  XIII.  —  Two  A  having  two  sides  and  the 
included  angle  of  the  one  equal  respectively  to  two  sides  and 
the  included  angle  of  the  other  are  congruent. 

The  data  are:  Two  A,  ABC  and  AB'C,  having  the 
three  equalities,  AB  =  A'B\  AC==A'C',  a  =  u'  (Fig.  27). 


Fig.  27. 


Proof.    Fit  the  angle  «  on  the  angle  a' ;     this  is  possible, 
because  the  angles  are  eqwal  and  congruent.     Then  A  falls 


36 


GEOMETRY. 


[Th.  XIV. 


on  A^ ',  also  the  point  B  falls  on  B^  (why?  Because  AB 
=  A'B^),  and  C  falls  on  C  (why?).  Hence  the  three  ver- 
tices of  the  two  A  coincide  in  pairs ;  therefore  the  three 
sides  of  the  two  A  coincide  in  pairs  (why?  Because  through 
two  points,  as  A  (A')  and  B  {B'),  only  one  ray  can  pass). 

Q.  E.  D. 

Corollary  i.  The  other  parts  of  the  two  A  are  equal 
or  congruent  in  pairs  of  correspondents  :  /?  =  /?',  7  =  y', 
BC=:B'C'. 

Corollary  2.  Pairs  of  equal  parts  lie  opposite  to  pairs  of 
equal  parts. 

48.  Theorem  XIV.  —  Two  A  having  two  angles  and  the 
included  side  of  the  one  equal  respectively  to  two  angles  and 
the  included  side  of  the  other  are  congruent  (Fig.  28). 


Fig.  28. 


Data:  Two  A  ABC,  A'B'C\  having  «=«',  ^  =  ^\ 
AB=A'B\ 

Proof.  Fit  AB  on  A'B';  this  is  possible  (M'hy?).  Then 
a  will  fit  on  a'  (why?),  and  ^  on  ft'  (why?)  ;  i.e.  the  ray 
^C  will  fit  on  A'C,  and  the  ray  BC  on  B'C  Then  the 
point  C  will  fall  on  C  (why  ?  Because  two  rays  meet  in  only 
one  point)  ;     i.e.  the  two  A  fit  exactly,     q.  e.  d. 


Th.  XVI.] 


TRIANGLES. 


37 


49.  We  m:iy  now  use  the  conditions  of  congruence  thus 
far  established  to  generate  new  notions  that  may  be  used  in 
estabHshing  other  Theorems. 

Dff.  The  ray  normal  to  a  tract  at  its  mid-point  is  called 
the  mid-normal  of  the  tract. 

Theorem  XV.  — Any  point  on  the  mid-normal  of  a  tract  is 
equidistant  from  its  ends   (Fig.  29). 


P>..C 


Data :  AB  a  tract,  M  its  mid-point,  Z  the  mid-normal, 
PsLny  point  on  it. 

Proof.  Compare  the  A  APM  and  BPM.  We  have 
AAf=BM  (why?).  PM=PM,  ^  AMP=:^  BMP 
(why?)  ;     hence  the  A  are  congruent  (why?)  ;    and  PA  = 

PB.      Q.  E.  D. 

Dff.  A  A  with  two  equal  sides,  like  APB,  is  called 
isosceles ;  the  third  side  is  called  the  base,  and  its  opposite 
angle  the  vertical  angle. 

50.  Theorem  XVI.  —  The  angles  at  the  base  of  an  isosceles 
A  are  equal;  and  conversely. 


38  GEOMETRY.  [Th.  XVI. 

Data:  ABC  an  isosceles  A,  AB  its  base,  AC  and  BC  its 
equal  sides  (Fig.  30). 


Proof.  Take  up  the  A  ABC,  turn  it  over,  and  replace  it 
in  the  position  BCA,  Then  the  two  A  ACB  and  BCA 
have  the  equal  vertical  angles,  C  and  C,  also  the  side  AC  = 
BC  (why?)  and  BC  =  AC  (why?)  ;  hence  they  are  con- 
gruent (why?),  and  the  '^A  =  :^B.     Q.  e.  d. 

Conversely,  A  A  whose  basal  angles  are  equal  is  isosceles. 
Let  the  student  conduct  a  proof  quite  similar  to  the  fore- 
going. 

Def.  The  ray  through  a  vertex  and  the  mid-point  of  the 
opposite  side  is  called  the  medial  of  that  side. 

Corollary  i .  In  an  isosceles  A  the  medial  of  the  base  is 
normal  to  it,  and  is  the  mid-ray  of  the  vertical  angle. 

Corollary  2.  When  the  medial  of  a  side  of  a  A  is  normal 
to  the  side,  the  A  is  isosceles.     Prove  it. 

Corollary  3.  When  the  medial  of  a  side  bisects  the 
opposite  angle,  the  A  is  isosceles.     Can  you  prove  it  ? 


Th.  XVI.]  LOGICAL   DIGRESSION.  39 


LOGICAL   DIGRESSION. 

51.  When  the  subject  and  predicate  of  a  proposition  are 
merely  exchanged,  the  proposition  is  said  to  be  converted^ 
and  the  new  proposition  is  called  the  converse.  Thus  X  is 
K;  coiiicrsch\  Kis  X.  In  general,  converses  of  true  prop- 
ositions are  not  true,  but  false.  Thus,  The  horse  is  an 
animal  is  always  correct,  but  The  animal  is  a  horse  is 
generally  false.  A  proposition  remains  true  after  simple 
conversion  only  when  subject  and  predicate  are  properly 
quantified,  thus  :  All  horses  are  some  animals  ;  conversely, 
Some  animals  are  all  horses.  Both  propositions  are  correct 
and  mean  the  same  thing.  But  they  are  awkward  in  ex- 
pression, and  such  forms  are  rarely  or  never  used.  When 
the  quantifying  word  is  all  or  its  equivalent,  the  term  is 
said  to  be  taken  universally  ;  when  it  is  some  or  its  equiva- 
lent, the  term  is  said  to  be  taken  particularly.  Thus  in  the 
foregoing  example  horse  is  taken  universally,  but  animal 
particularly.  The  only  useful  conversions  are  of  proposi- 
tions in  which  both  subject  and  predicate  are  ufiiversal.  In 
the  great  body  of  propositions  only  the  subject  is  quantified 
universally,  the  quantifier  is  omitted  from  the  predicate,  but 
a  particular  one  is  understood.  To  show  that  a  universal 
quantifier  is  admissible  requires  in  general  a  distinct  proof. 

52.  In  order  to  convert  an  hypothetic  proposition,  we 
exchange  hypothesis  and  conclusion.  Thus,  \{X  is  K,  U  is  F; 
the  converse  is,  if  6^"  is  V^  X  is  K  All  such  hypothetic 
propositions  may  be  stated  categorically ^  thus  :  All  cases  of 
A' being  Fare  cases  of  6^"  being  V  \  conversely,  All  cases  of 
U  being  V  are  cases  of  X  being  Y.  This  converse  is  plainly 
false  except  when  the  quantifier  all  is  admissible  in  the  first 
predicate. 


40  GEOMETRY.  [Th.  XVII. 

53.  But  while  the  converse  of  a  true  hypothetic  propo- 
sition is  generally  false,  the  contrapositive  is  always  true. 
This  latter  is  formed  by  exchanging  hypothesis  and  conclu- 
sion and  denying  both.  Thus  :  If  Jf  is  K,  then  17  is  F; 
contrapositive y  If  U  is  not  V,  then  X  is  not  K  Or,  if  a 
point  is  on  the  mid-normal  of  a  tract,  then  it  is  equidistant 
from  the  ends  of  the  tract ;  contrapositive.  If  a  point  is  not 
equidistant  from  the  ends  of  a  tract,  then  it  is  not  on  the 
mid-normal  of  the  tract. 


54.  Theorem  XVII.  — An  outer  angle  of  a  A  is  greater 
than  either  inner  non-adjacent  angle. 

Data:  Let  ABC  be  any  A,  a'  an  outer  angle,  ^'  a  non- 
adjacent  inner  one  (Fig.  31). 


Fig.  31. 

Proof.  Draw  the  medial  CM  and  lay  off  MD  —  MC ; 
also  draw  AD.  Then  in  the  A  AMD  and  BMC  we  have 
AM=BM  {vfhy}),  MD  =  MC  (why?),  and  '^AMD  = 
'^.BMC  (why?)  ;  hence  the  A  are  congruent  (why?),  and 
^MBC=^MAD  (why?).  But  -^^  MAD  is  only  part  of 
the  ^  «' ;  hence  «'>  ^  MAD  (why?)  ;  i.e.  «'>/?'.     q.  e.  d. 

Similarly,  prove  that  a!  >  y. 


Th.  XVIII]  triangles,  41 

55.  Theorem  XVIII.  — If  tivo  sides  0/  a  A  are  unequal, 
then  the  opposite  angles  are  unequal  in  the  same  sense  {i.e. 
the  greater  angle  opposite  the  greater  side)  (Fig.  32). 


Fig.  32. 

Data:  ABC  a  A,  AC  >  AB,  AR  the  mid-ray  of  the 
angle  at  A,  AB'  laid  off  =  AB. 

Proof.  ABR  and  AB^R  are  congruent  (why?)  ;  hence 
^  ABR  =  ^  AB'R  (why  ?)  ;  but  ^  AB'R  >  C  (why  ?)  ; 
i.e.  7(.ABC>^ACB.     q. e. d. 

Conversely,  If  two  angles  0/  a  A  are  unequal,  the  opposite 
sides  are  unequal  in  the  same  sense. 

Proof.  The  opposite  sides  are  not  equal ;  for  when  the 
sides  are  equal,  the  opposite  angles  are  equal  (Theorem 
XVI.),  and  contrapositively,  when  the  angles  are  unequal, 
the  opposite  sides  are  unequal.  Then,  by  the  preceding 
Theorem,  the  greater  angle  lies  opposite  the  greater  side. 

56.  Join  BB' ;  then  AR  is  the  mid-normal  of  BB'  (why  ?), 
and  hence  angle  CBB'=  angle  BB^R  (why  ?) .  Hence  angle 
BB'C>B'BC  (why?)  ;  hence  BC>B'C  (why?).  But 
B'C=AC-AB;  hence  BC>  AC  -  AB  :  i.e. 


42 


GEOMETRY. 


[Th.  XIX. 


Theorem  XIX.  —  Any  side  of  a  ^  is  greater  than  the 
difference  of  the  other  two. 

Add  AB  to  both  sides  of  this  inequahty  and  there  results 
AB  +  BOAC]  i.e. 

Theorem  XX.  —  Any  side  of  a  A  is  less  than  the  sum  of 
the  other  two. 

This  fundamental  Theorem  is  here  proved  on  the  sup- 
position that  AB  <AC;  if  AB  were  =ACot>  AC, it  would 
need  no  formal  proof. 

57.  Theorem  XXI.  — A  point  not  on  the  mid-normal  of 
a  tract  is  not  equidistant  from  the  ends  of  the  tract. 

Data :  AB  the  tract,  MN  the  mid-normal,  Q  any  point 
not  oviMN  {Y\g.  33). 


Fig.  33. 

Proof.  Draw  QA  and  QB  \  one  of  them,  as  QA,  must 
cut  MN^X.  some  point,  as  P.  Then  QB<  QP-\-PB  (why  ?) , 
and  PB^PA  (why?);  hence  QB<QP-\-PA',  i.e. 
QB  <  QA.     Q.  E.  D. 

Of  what  Theorem  is  this  the  converse  ? 

If  now  we  seek  for  a  point  equidistant  from  A  and  B,  we 
can  find  it  on  the  mid-normal  of  AB  and  only  there  ;  hence 
the  locus  of  a  point  equidistant  from  the  ends  of  a  tract  is 
the  mid- normal  of  the  tract. 


Th.  XXIII.] 


TRIANGLES, 


43 


58.  Theorem  XXII.  —  Two  A  with  the  three  sides  of  the  one 
equal  respectively  to  the  three  sides  of  the  other  are  congruent. 

Data:  ABC  and  A'B'C  the  two  A,  and  AB  =  AB\ 
BC==B'C,    C^=  C'y^' (Fig.  34). 


Proof.  Turn  the  A  A'B'C  over  and  fit  A'B^  on  AB  so 
that  C  shall  fall  (say)  at  D.  Draw  CD.  Then  A  and  B 
are  on  the  mid-normal  of  CD  (why?)  ;  hence  the  ray  AB 
is  the  mid-normal  of  CD  (why?)  ;  hence  the  angle  CAB  = 
angle  DAB,  and  angle  CBA  =  angle  DBA  (why?).  Hence 
the  A  are  congruent  (why?),     q.  e.  d. 

N.B.  As  to  when  the  A  must  be  turned  round  and  when 
turned  over,  see  Art.  94. 

59.  Theorem  XXIII.  —  A.  From  any  point  outside  of  a 
ray  one  normal  may  be  drawn  to  the  ray. 

Data:    /*the  point,  LV  the  ray  (Fig.  35). 

Proof.  From  P  draw  a  ray  far  to  the  left,  as  PA^  making 
the  angle  PAL  >  angle  PAL\  Now  let  the  ray  turn  about 
/*  as  a  pivot  into  some  position  far  to  the  right,  making 
angle  PA^L  <  PA^V.  The  plane,  the  angle,  the  motion,  all 
being  continuous,  in  passing  from  the  stage  of  being  unequal 


44 


GEOMETRY. 


[Tii.  XXIII. 


in  one  sense  to  the  stage  of  being  unequal  in  the  opposite 
sense,  the  angles  made  by  the  moving  ray  with  the  fixed  ray 
must  have  passed  through  the  stage  of  equality.     Let  PN  be 


Fig.  35. 

the  ray  in  this  position  so  that  angle  PNL  —  angle  PNL!  ; 
then  each  is  a  right  angle  by  Definition,  and  /Wis  normal  to 

LL.      Q.  E.  D. 

B.  There  is  only  one  ray  through  a  fixed  point  and  normal 
to  a  fixed  ray. 

Proof.  Any  other  ray  than  PN,  as  PD,  is  not  normal  to 
LV ;  for  the  outer  angle  PDL  is  >  the  right  angle  PND 
(why?).     Q.  E.  D. 

C.  The  normal  tract  PN  is  shorter  than  any  other  tract 
from  P  to  the  ray  LV . 

Proof.  For  the  right  angle  at  Wis  >  angle  PDN  (why?)  ; 
hence  PN  <  PD  (why  ?) .     q.  e:.  d. 

D.  E.  Equal  tracts  from  point  to  ray  meet  the  ray  at 
equal  distances  from  the  foot  of  the  normal;  and  conversely. 

Proof.  For,  if  DPD^  be  isosceles,  then  the  normal  PN  is 
the  medial  of  the  base  (why  ?) . 

F.  Two,  and  only  t7i>o,  tracts  of  given  length  can  be  drawn 
from  a  point  to  a  ray. 


Th.   XXIV.] 


TRIANGLES. 


45 


Proof.  For  two,  and  only  two,  points  are  on  the  ray  at  a 
given  distance  from  the  foot  of  the  normal. 

G.  Of  tracts  drawn  to  points  unequally  distant  from  the 
foot  of  the  normaly  the  one  drawn  to  the  remotest  is  the 
longest. 

Proof.  In  the  A  PDA,  angle  PDA  >  PAD  (why?)  ; 
hence  PA  >  PD  (why  ?) .     q.  e.  d. 

Similarly,  PA'  >  PD. 

H.  Equal  tracts  from  the  point  to  the  ray  make  equal 
angles  with  the  normal  from  the  point  to  the  ray  and  also 
equal  angles  with  the  ray  itself;  and  conversely. 

I.  Of  unequal  tracts  from  the  point  to  the  ray,  the  longest 
makes  the  greatest  angle  with  the  normal  and  the  least  with 
the  ray. 

I^t  the  student  conduct  the  proof  of  H  and  /. 

60.  Theorem  XXIV.  —  Two  A  having  two  angles  and  an 
opposite  side  of  one  equal  respectively  to  two  angles  and  an 
opposite  side  of  the  other  are  congruent. 

Data  :  ABC  and  A^B'C  two  A  having  AB  —  A'B\  angle 
u  =  angle  a',  angle  y  =  angle  y'  (Fig.  36). 


Fig.  36. 


46 


GEOMETRY. 


[Th.  XXIV. 


Proof.  Fit  «'  on  a  ;  then  B'  falls  on  B  (why?),  and  AC 
falls  along  A  C.  Draw  the  normal  BN.  Then  ^  C  and  B  C 
make  the  same  angle,  y  =  y',  with  the  ray  AN;  hence  they 
are  =  and  meet  the  ray  in  the  same  point  (why  ?)  ;  i.e.  C 
falls  on  C ;   i.e.  the  A  are  congruent.     Q.  e.  d. 

6i.  We  now  come  to  the  so-called  ambiguous  case,  of 
two  A  with  two  sides  and  an  opposite  angle  in  one  equal  to 
the  two  sides  and  the  corresponding  opposite  angle  in  the 


Fig.  37. 

other.    Let  ABC  and  A B' C  (Fig.  37)  be  the  two  A,  with 
AB  =  A'B',  BC=B'C,  and  angle  a  =  angle  «'.     Fit  a'  on 


Th.  XXV.] 


TRIANGLES. 


47 


«  ;  then  A^B^  falls  on  ABy  B^  on  B ;  but  since  from  a  point 
B  {B')  we  may  draw  two  equal  tracts  to  the  ray  AL,  the 
side  B'C  may  be  either  of  these  equals  and  may  or  may  not 
fall  on  BC,  In  general,  then,  we  cannot  prove  congnience 
in  this  case.  But  if  ^C  be  >  AB^  then  angle  «  >  angle  y 
(why?),  and  there  is  only  one  tract  on  the  right  of  AB  drawn 
from  B  to  the  ray  AC  and  equal  io  BC ;  the  other  tract 
equal  to  BC  must  be  drawn  outside  of  AB  and  to  the  left. 
Hence  in  this  case,  when  the  angle  lies  opposite  the  greater 
side,  the  A  are  congruent.     Hence 

Theorem  XXV.  —  Two  A  having  two  sides  and  an  angle 
opposite  the  greater  in  one  equal  to  two  sides  and  an  angle 
opposite  the  greater  side  in  the  other  are  congruent. 

Corollary.  Two  right  A  having  a  side  and  any  other 
part  of  one  equal  to  a  side  and  the  corresponding  part  of 
the  other  are  congruent. 


Fig.  38. 


62.    We  have  seen  (Art.  47)  that  when  two  A  have  two 
sides  and  included  angle    in  one  equal  to  two  sides  and 


48  GEOMETRY.  [Tii.  XXVI. 

included  angle  in  the  other,  they  are  congruent.  But  what 
if  the  included  angles  are  not  equal  ?  Let  ABC  and  A'B' C 
be  the  two  A,  having  AB^AB\  BC=  B'C,  but  y8  >  /8'. 
Slip  the  upper  film  of  the  plane  along  until  A^B^  fits  on  AB 
and  let  C  fall  on  D.  Draw  the  mid-ray  BM  of  the  angle 
CBD,  let  it  cut  AC  Tui  M,  and  draw  DM.  Then  the  A 
CBM SiXid  DBM 3ixe  congruent  (why?)  ;  hence  AM-{-  MD 
=  AC  (why?),  and^C>^Z>,  or^O^'C.     Hence 

Theorem  XXVI.  —  Two  A  having  two  sides  of  one  equal 
to  two  sides  of  the  other,  but  the  included  angles  unequal^ 
have  also  the  third  sides  unequal,  the  greater  side  lying 
opposite  the  greater  angle. 

Conversely,  Two  A  having  two  sides  in  one  equal  to  two 
sides  in  the  other,  but  the  third  sides  unequal,  have  the 
included  angles  also  unequal,  the  greater  angle  being  opposite 
the  greater  side. 

Proof.  The  included  angles  are  not  equal;  for  if  they 
were  equal,  the  A  would  be  congruent  (why?)  and  the 
three  sides  would  be  equal.  Hence  the  included  angles  are 
unequal,  and  the  relation  just  estabUshed  holds  ;  namely,  the 
greater  angle  Hes  opposite  a  greater  side.     q.  e.  d. 

63.  Theorem  XXVII. — Every  point  on  a  mid-ray  of  an 
angle  is  equidistant  from  its  sides. 

Data :    O  the  angle,  MJVV  the  mid-ray,  P  any  point  on  it. 

Proof.  From  P  draw  the  normals  PC  and  PD  ;  they  are 
(Fig.  39)  the  distances  of  P  from  the  ends  of  the  angle. 
Then  the  A POC  and  POD  are  congruent  (why?)  ;     hence 

PD  =  PC.      Q.  E.  D. 


Th.  XXVIII.] 


TRIANGLES. 


49 


Conversely,  A  poitit  equidistant  from  the  ends  of  an  angle 
is  on  a  mid-ray  of  the  angle  (Fig.  39). 


FIG.  39. 

Proof.  If  PC  =  PD,  then  the  ^  POC  and  POD  are 
congruent  (why  ?)  ;  hence  angle  POD  =  angle  POC.   q.  e.  d. 

Accordingly  we  say  that  the  mid-rays  of  an  angle  are  the 
locus  of  a  point  equidistant  from  its  ends. 

*64.  It  is  just  at  this  stage  in  the  development  of  the 
doctrine  of  the  Triangle  that  we  are  compelled  to  halt  and 
introduce  a  new  concept  before  we  can  proceed  any  further. 
The  necessity  of  this  step  will  appear  from  what  follows 
(which  may,  however,  be  omitted  on  first  reading,  at  the 
option  of  teacher  or  student) . 

Def.  Two  A  not  congruent  are  called  equivalent  when 
they  may  be  cut  up  into  parts  that  are  congruent  in  pairs. 


Theorem  XXVIII.  —  Any  A  is  equivalent  to  another  A 
having  the  sum  of  two  of  its  angles  equal  to  the  smallest 
angle  of  the  given  A. 

Data  :  ABC  the  A,  a  the  least  angle  (Fig.  40). 


50  GEOMETRY.  [Th.  XXIX. 

Proof.  Through  M,  the  mid-point  of  BC,  draw  AM  SLud 
make  MD  =  MA.  Then  the  A  A  CM  and  DBM  are  con- 
gruent (why?),  the  part  AMB  is  common  to  ABC  and 
ABD,  and  the  sum  of  the  angles  ADB  and  ^^Z?  =  angle 

BAC.      Q.E.D. 


Fig.  40. 

Corollary.     The  sum  of  the  angles  in  the  new  A  is  equal 
to  the  sum  of  the  angles  in  the  old  A. 

*  65.  We  may  now  repeat  this  process,  applying  it  to  the 
smallest  angle,  as  A,  of  the  A  ABD.  In  the  new  A  ABE 
the  smallest  angle,  as  A,  cannot  be  greater  than  \  of  the 
original  angle  a  in  ABC ;  after  n  repetitions  of  this  process 
we  obtain  a  A,  as  ALB,  in  which  the  sum  of  the  angles  A 

and   L   cannot  be  >  —   of  the  original  angle  «  in  the  A 

^"  I 

ABC.     By  making  n  as  large  as  we  please,  we  make  — 

as  small   as  we    please,  and   so  we   make   —   of  angle  a 

smaller  than  any  assigned  magnitude  no  matter  how  small. 
Meantime  the  other  angle  B  has  indeed  grown  larger  and 
larger,  but  has  remained  <  a  straight  angle.  Hence  the 
sum  of  the  angles  in  the  A  ALB  cannot  exceed  a  straight 
angle  by  any  amount  however  small ;  but  the  sum  of  the 
angles  in  ALB  =  sum  of  the  angles  in  ABC ;  hence 

Theorem  XXIX.  —  The  sum  of  the  angles  in  any  A  can- 
not exceed  a  straight  angle  by  any  finite  amount. 


Th.  XXX.] 


TRIANGLES. 


51 


Corollary  i.  The  outer  angle  of  a  A  is  not  less  than  the 
sum  of  the  inner  non-adjacent  angles. 

Corollary  2.  From  any  point  outside  of  a  ray  there  may 
be  drawn  a  ray  making  with  the  given  ray  an  angle  small  at 
will. 

Proof.  From  P  draw  any  ray  PA,  and  lay  off  AB  =  PA 
(Fig.  41).    Then  the  angle    PBA    is    not   greater   than 


Fio.  41. 

\PAN  (why?);  now  lay  off  BC=^PB  (why?);  then 
angle  PCB  is  not  >  \  angle  PBA  (why?)  ;  proceeding 
this  way,  we   obtain  after  n  constructions  an  angle  PLN 

not  >  —  of  the  angle  PAN^  and  by  making  n  large  enough 
2" 

we  may  make  this  ^  as  small  as  we  please,    q.  e.  d. 


♦66.  Theorem  XXX.  —  If  the  sum  of  the  angles  in  any  A 
equals  a  straight  angle,  then  it  equals  a  straight  angle  in 
every  A  (Fig.  42). 


52 


GEOMETRY, 


[Th.  XXX. 


Hypothesis :    ABC  a   A   with   the    sum   of  its    angles 

Proof,  (i)  Draw  any  ray  through  C,  as  CD.  Then  if 
the  sum  of  the  angles  in  the  A  ACD  and  BCD  be  ^  — ;c 
and  S—y,  x  and  y  being  any  definite  magnitudes  however 
small,  then  on  adding  these  sums  we  get  2S—{x-\-y)', 
and  on  subtracting  the  sum,  S,  of  the  supplemental  angles 
at  Z>  we  get  6*  — (^-f-jv)  for  the  sum  of  the  angles  of  the 
A  ABC.  Now  if  this  sum  be  S,  then  x  and  y  must  each 
be  O ;  i.e.  the  sum  of  the  angles  in  each  of  the  A  A  CD 
and  BCD  is  5.  Now  draw  DE  and  DF\  in  each  of  the  four 
small  A  the  sum  of  the  angles  is  still  =  ^.  (2)  We  may 
now  make  a  A  as  large  as  we  please  and  of  any  shape  what- 
ever, but  the  sum  of  the  angles  will  remain  =  S.  For,  take 
the  same  A  ABC,  and  draw  CD  normal  to  AB.  Then  the 
sum  of  the   (Fig.  43)  angles  in  the  A  ACD  is  S,  as  has 


Fig.  43. 

been  shown  above  ;  also  angle  Z>  is  a  right  angle  ;  hence 
the  angles  A  and  ACD  are  complementary.  Now  along 
^C  fit  another  A  ACD^  congruent  with  ACD)  then  all 
the  angles  of  the  quadrilateral  AD  CD'  are  right,  and  the 
figure  is  called  a  rectangle.  Now  we  can  place  horizontally 
side  by  side  as  many  of  these  rectangles,  all  congruent,  as  we 
please,  say  p  of  them ;  we  can  also  place  as  many  of  them 
vertically,  one  upon  another,  as  we  please,  say  q  of  them  ; 


Th.  XXX.]  TRIANGLES.  53 

and  we  can  then  fill  up  the  whole  figure  into  a  new  rectan- 
gle, as  large  as  we  please.  About  each  inner  junction-point 
of  the  sides  of  the  rectangles  there  will  be  four  right  angles 
plainly.  Now  connect  the  two  opposite  vertices,  as  A  and  Z, 
of  this  rectangle.  So  we  get  two  congruent  right  A,  in  each 
of  which  the  sum  of  the  angles  is  S.  Then  any  A  that  we 
cut  off  from  this  right  A  will,  by  the  foregoing,  have  the 
sum  of  its  angles  equal  to  6".  Since  /  and  q  are  entirely  in 
our  power,  we  may  make  in  this  way  any  desired  right  A 
and  from  it  cut  off  any  desired  oblique  A,  with  the  sum  of 
its  angles  =  S.     q.  e.  d. 

Hence  either  no  A  has  the  sum  of  its  angles  =  6",  or 
every  A  has  the  sum  of  its  angles  =  S. 

67.  A  logical  choice  between  these  alternatives  is  impos- 
sible, but  the  matter  may  be  cleared  up  by  the  following 
considerations : 

Across  any  ray  LM  draw  a  transversal  T,  cutting  LM  zX 
O,  and  making  the  angles  «,  )3,  y,  8.  Through  any  point, 
as  O',  of  T'draw  a  ray  (Fig.  44)  L'M'  making  angle  «'  =  a. 


Fig.  44. 

This  is  evidently  possible  (why?).      Then  plainly  /8'  =  /3, 
y'  =  y,  S'  =  8,  «'  =  u  ;  they  are  called  corresponding  angles  ; 


5+  GEOMETRY.  [Th.  XXX. 

also  a  and  y',  /?  and  V  are  equal,  —  they  are  called  alternate 
angles ;  also  «  and  V,  as  well  as  ^  and  y',  are  supplemental, 
—  they  are  called  interadjacent  angles. 

68.  Now  let  /*be  the  mid-point  oi  00^ ;  on  it  as  a  pivot 
turn  the  whole  right  side  of  the  plane  round  through  a 
straight  angle  until  O  falls  on  0\  and  6>'  falls  on  O.  Then, 
since  the  angles  about  O  and  O^  are  equal  as  above  stated, 
the  half-ray  OL  will  ffill  and  fit  on  the  half-ray  0'M\  and 
the  half-ray  O^V  on  the  half-ray  OM.  Accordingly,  if  the 
rays  LM  and  Z'J/'  meet  on  one  side  of  the  transversal  T, 
they  also  meet  on  the  other  side  of  T. 

69.  Three  possibilities  here  lie  open  : 

(i)  The  rays  ZJ/and  Z'J/'  may  meet  on  the  left  and 
also  on  the  right  of  T,  in  different  points. 

(2)  They  may  meet  on  the  left  and  also  on  the  right  of 
T,  in  the  same  point. 

(3)  They  may  not  meet  at  all. 

No  logical  choice  among  these  three  is  possible.  But 
in  all  regions  accessible  to  our  experience  the  rays  neither 
converge  nor  show  any  tendency  to  converge.  Hence  we 
assume  as  an 

Axiom  A.  Two  rays  that  make  with  any  third  ray  a  pair 
of  corresponding  angles  equal,  or  a  pair  of  alternate  angles 
equal,  or  a  pair  of  interadjacent  angles  supplemental,  are 
non-intersectors. 

70.  But  another  query  now  arises.  Is  it  possible  to  draw 
another  ray  through  (9'  so  close  to  Z'  that  it  will  not  meet 
'OL  however  far  both  may  be  produced  ?  Here  again  it  is 
impossible  to  answer  from  pure  logic.  An  appeal  to  experi- 
ence is  all  that  is  left  us.     This  latter  testifies  that  no  ray 


Th.  XXX.]  PARALLELS,  55 

can  be  drawn  through  O^  so  close  to  0*V  as  not  to  approach 
and  finally  meet  the  ray  OL.     Hence  we  assume  as  another 

Axiom  B.  Through  any  point  in  a  plane  only  one  non- 
intersector  can  be  drawn  for  a  given  straight  line. 

This  single  non-intersector  is  commonly  called  the  parallel, 
through  the  point,  to  the  straight  line. 

71.  It  cannot  be  too  firmly  insisted,  nor  too  distinctly 
understood,  that  the  existence  of  any  non-intersector  at  all, 
and  the  existence  of  only  one  for  any  given  point  and  given 
ray,  are  both  assumptions,  which  cannot  be  proved  to  be 
facts.  The  best  that  can  be  said  of  them,  and  that  is  quite 
good  enough,  is  that  they  and  all  their  logical  consequences 
accord  completely  and  perfectly  with  all  our  experience  as 
far  as  our  experience  has  hitherto  gone.  Even  then,  if 
there  be  any  error  in  our  assumptions,  we  have  thus  far  been 
utterly  unable  to  find  it  out. 

A  geometry  that  should  reject  either  or  both  of  these 
assumptions  would  have  just  as  much  logical  right  to  be  as 
the  geometry  that  accepts  them,  and  such  geometries  lack 
neither  interest  nor  importance.  They  may  be  called  Hyper- 
Euclidean  in  contradistinction  from  this  of  ours,  which  from 
this  point  on  is  Euclidean  (so-called  from  the  Greek  master, 
Euclides,  who  distinctly  enunciated  the  equivalent  of  our 
Axioms  in  a  Definition  and  a  Postulate). 

Note.  —  Observe  the  relation  of  Axioms  A  and  B :  the  one  is  the 
converse  of  the  other. 

Observe  also  that  the  necessity  of  assuming  the  first  lies  in  our  igno- 
rance of  the  indefinitely  greats  and  the  occasion  of  assuming  the  other 
lies  in  our  ignorance  of  the  indefinitely  small.     See  Note,  Art.  301. 

73.  Accepting  our  Axioms  as  at  least  exacter  than  any 
experiment  we  can  makc^  we  may  now  easily  settle  the  ques- 


56  GEOMETRY.  [Th.  XXXI. 

tion  as  to  the  sum  of  the  angles  in  a  A.  Let  ABC  be  any 
A ;  through  the  vertex  C  draw  the  one  parallel  to  the  base 
AB.  Then  a  =  «',  ^  =  /S'  (why?)  ;  also  a'  +  y  +  )8'  =  6" j 
hence  a  +  y-|-/3  =  ^/  i.e.  (Fig.  45) 

Theorem  XXXI.  —  The  sui7i  of  the  angles  in  a  A  is  a 
straight  angle. 


Fig.  4S. 

Corollary  i.  The  outer  angle  E  equals  the  sum  of  the 
inner  non-adjacent  angles  a  and  y  (why?). 

Corollary  2.  If  two  angles  of  a  A  be  known,  the  third  is 
also  known. 

Corollary  3.  If  two  A  have  two  angles,  or  the  sum  of  two 
angles  of  the  one  equal  to  two  angles,  or  the  sum  of  two 
angles  of  the  other,  then  the  third  angles  are  equal. 

^Corollary  4.  To  know  the  three  angles  of  a  A  is  not  to 
know  the  A  completely,  for  many  A  may  have  the  same 
three  angles.  Such  A  are  similar,  as  we  shall  see,  but  are 
not  congruent;  they  are  alike  in  shape,  but  not  in  size. 

73.  Next  to  normality,  parallelism  is  the  most  important 
relation  in  which  rays  can  stand  to  each  other,  and  we  must 
now  use  the  new  relation  in  the  generation  of  new  concepts. 


Til.  XXXIII  ] 


PARALLELOGRAMS. 


57 


Theorem  XXXII.  — Parallel  Intercepts  between  parallels 

arc  e<]ual. 

Data :   L  and  Z',  it/and  M ,  two  pairs  of  parallels  (Fig.  46). 


Fig.  46. 

Proof.  Draw  BD.  Then  the  A  ABD  and  CDB  are 
congruent  (why?),  and  AB  =  CD,  BC=DA.    q.  e. d. 

Def.  The  figure  ABCD  formed  by  two  pairs  of  parallel 
sides  is  called  a  parallelogram,  and  may  be  denoted  by  the 
symbol  O. 

A  join  of  opposite  vertices,  as  BD,  is  called  a  diagonal. 

74.    Theorem  XXXIII.  —  Properties  of  the  parallelogram. 

A.  The  opposite  sides  of  a  parallelogram  are  equal. 
This  has  just  been  proved. 

B.  The  opposite  angles  of  a  parallelogram  are  equal. 
Proof.     «  =  /3  (why?);   )8  =  «'   (why?);   hence  « = «'. 

Q.  E.  D. 

Corollary,  Adjacent  angles  of  a  parallelogram  are  sup- 
plementary. 

C.  Each  diagonal  of  a  parallelogram  cuts  it  into  two  con- 
gruent A.     Prove  it. 


58 


GEOMETRY. 


[Th.  XXXIV. 


D.    The  diagonals  of  a  parallelogram  bisect  each  other 
(Fig.  47) 


,-''M 


FIG.  47. 

Proof.     The  A  AMB  and  CMD  are  congruent  (why?)  ; 
hence  AM^  CM,  BM=  DM.     q.  e.  d. 

75.   We  may  now  convert  all  the  foregoing  propositions 
and  obtain  as  many  criteria  of  the  parallelogram. 

Theorem  XXXIV.  —  A'.   A  4- side  with  its  opposite  sides 
equal  is  a  parallelogram. 

Data  :    AB  ^  CD,  AD=  CB  (Fig  48). 


W^-^, 


Fig.  48. 

Proof.  Draw  BD.  Then  ABD  and  CDB  are  congruent 
(why?);  hence  ^  =  8;  and  AD  and  CB  are  parallel; 
similarly,  AB  and  CD  are  parallel ;  hence  ABCD  is  a  par- 
allelogram.    Q.  E.  D. 


Th.  XXXIV.]  PARALLELOGRAMS.  59 

B'.   A  4-side  with  opposite  ant^ks  equal  is  a  parailelogram. 
Data:    «  =  «',  ^  +  yS' =  y  4- y'  (Fig.  49). 


'i-A 


Fig.  49. 

Proof.  Since  «  =  «',  y3  +  y  =  )3'  +  y'  (why?).  Hence 
/?  =  y',  ft'  =  y;  i.e.  opposite  sides  are  parallel,  the  4-side 
is  a  parallelogram,     q.  e.  d. 

C,  A  4-side  that  is  cut  by  each  diagonal  into  two  congru- 
ent A  is  a  parallelogram. 

For  the  opposite  angles  must  be  equal  (why?)  ;  hence, 
etc.     Q.  E.  D. 

D'.  A  4-side  whose  diagonals  bisect  each  other  is  a 
parallelogram. 

For  the  opposite  sides  are  equal,  being  opposite  equal 
angles  in  congruent  A  ;  hence,  etc.     q.  e.  d. 

E'.  A  4-side  with  one  pair  0/  sides  equal  and  parallel  is 
a  parallelogram. 

For  the  other  two  sides  are  equal  and  parallel  (why?)  ; 
hence,  etc.    q.  e.  d. 

76.  The  foregoing  properties  and  criteria  of  the  parallel- 
ogram illustrate  excellently  the  nature  of  a  definition.     This 


60  GEOMETRY.  [Th.  XXXV. 

latter  defines  or  bounds  off  by  stating  something  that  is  true 
of  the  thing  defined,  but  of  nothing  else.  Accordingly,  the 
characteristic  of  every  definition  or  definitive  property  is 
that  the  proposition  that  states  it  may  be  converted  simply. 
Thus: 

Every  parallelogram  is  a  4-side  with  opposite  angles 
equal ;  and  conversely,  every  4-side  with  opposite  angles 
equal  is  a  parallelogram. 

Not  every  property  is  definitive,  and  hence  not  every 
property  may  be  used  as  test  or  criterion. 

77.    Special  Parallelograms. 

Def.   An  equilateral  parallelogram  is  called  a  rhombus. 

Theorem  XXXV.  —  The  diagonals  of  a  7'hombus  a7'e  nor- 
mal to  each  other. 

Let  the  student  conduct  the  proof  suggested  by  the  figure 
(Fig-  50)- 


Fig.  50. 

Conversely,  A  parallelogram  ivhose  diagonals  are  normal 
to  each  other  is  equilateral,  or  a  rhombus.  Let  the  student 
supply  the  proof. 

78.  Def.  An  equiangular  parallelogram  is  called  a  rect- 
angle (for  all  the  angles  are  right  angles) . 


Th.  XXXVII.]  PARALLELOGRAMS. 


61 


Theorem  XXXVI.  —  The  diagonals  of  a  rectangle  are  equal 
(fig-  50- 


D 

C 

A 

B 

Fig.  51. 
For  the  A  ABC  and  BAD  are  congruent  (why?)  ;  hence 

AC=BD.      Q.E.D. 

Conversely,  A  parallelogram  with  equal  diagonals  is  equi- 
angular^  or  a  rectangle. 

For  the  A  ABC  and  BAD  are  again  congruent,  though 
for  another  reason.     What  reason  ? 

79.  Def.  A  parallelogram  both  equilateral  and  equi- 
angular is  called  a  square. 

Theorem  XXXVII.  —  The  diagonals  of  a  square  are  equal 
and  normal  to  each  other. 


Fio.  52. 


62  GEOMETRY.  [Th.  XXXVIII. 

For  the  square,  being  both  rhombus  and  rectangle,  has 
all  the  definitive  properties  of  both.  Or  the  student  may 
prove  the  proposition  directly  from  the  figure  (Fig.  52),  as 
well  as  its  converse  : 

A  parallelogram  with  diagonals  equal  and  normal  to  each 
other  is  a  square. 

80.  Can  we  convert  Theorem  XXXII.  and  prove  that 
equal  intercepts  between  parallels  are  parallel  ?  Manifestly 
no  (Fig.  53),  for  from  the  point  C  we  may  draw  two  equal 


Fig.  53. 

tracts  to  the  other  parallel,  the  one  CB  parallel  to  AD,  the 
other  CB^  sloped  at  the  same  angle  to  the  parallels  but  in 
opposite  ways.  We  may  call  CB^  a?iti-parallel  to  AD,  and 
the  figure  AB'CD  an  anti-parallelogram.  Since  from  any 
point  C  only  two  equal  tracts,  or  tracts  of  given  length,  may 
be  drawn  to  the  other  parallel  through  A,  we  have  the 

Theorem  XXXVIII.  —  Equal  intercepts  between  parallels 
are  either  parallel  or  anti-parallel. 

Corollary  i.  Adjacent  angles  of  an  anti-parallelogram 
are  alternately  equal  or  supplemental. 

Corollary  2.  Anti-parallels  prolonged  meet  at  the  vertex 
of  an  isosceles  A. 


Th.  XL.] 


GENERAL   QUADRILATERAL. 


63 


THE   GENERAL   QUADRILATERAL   OR   4-SIDE. 

81.  A  Quadrilateral  is  determined  by  four  intersecting 
rays.  These  determine  six  points,  the  four  inner  vertices, 
C,  Z>,  Ey  Fj  and  the  two  outer  ones,  Aj  B.  The  cross-rays, 
CE^  DFy  AB,  are  the  diagonals,  CE  and  DF  itiner,  AB 
outer.  Commonly  the  outer  diagonal  is  little  used,  and  the 
inner  ones  are  called  the  diagonals.  When  none  of  the 
angles  C,  Z>,  E^  F,  of  the  4-side  is  greater  than  a  straight 
angle,  the  4-side  is  called  the  normal,  as  CDEF.  It  is  the 
only  form  ordinarily  considered.  The  other  two  forms  are 
(2)  the  crossed^  ACBE^  and  (3)  the  inverse^  ADBF 
(Fig.  54).     For  all  forms  let  the  student  prove 


Fio.  54. 

Theorem  XXXIX.  —  The  sum  of  the  inner  angles  of  a 
^-side  is  a  round  angle. 

Corollary.     When  two  angles   of  a   4-side  are   supple- 
mental, so  are  the  other  two. 

82 .    Theorem  XL.  —  The  angles  between  two  rays  equal 
the  angles  between  two  normals  to  the  rays. 


64 


GEOMETRY. 


[Th.  XLI. 


Data :    OL  and  OM  any  two  rays,  PA  and  PB  any  two 
normals  to  them  (Fig.  55). 


/ 

\ 

/3' 

P 

r/^ 

B 

A 

M 

Fig.  55. 

Proof.  The  angles  at  A  and  B  are  right  angles  and 
therefore  supplemental  (why?)  ;    hence  a  —  «',  and  p  =  yS'. 

Q.  E.  D. 

N.B.  The  4-side  with  its  opposite  angles  supplemental  is 
very  important  and  has  received  the  name  encyclic  4-side, 
for  reasons  to  be  seen  later  on  (Arts.  126-7). 


THREE   OR   MORE    PARALLELS. 

83.  Theorem  XLI.  —  Three  parallels  that  make  equal 
intercepts  on  one  transversal,  make  equal  intercepts  ofi  any 
transversal. 

Data  :  Z,  M,  N,  three  parallels,  and  AB  =  BCy  and  DEF 
any  transversal  (Fig.  56). 

Proof.  Draw  D'EP  parallel  to  AB C.  Then  AB  =  BC 
(why?),  AB  =  D^E  (why?),  and  BC=^EF  (why?); 
hence  D^E  =  EF  (why?),  hence  the  A  BED'  and  FEF 
are  congruent  (why?)  ;    hence  DE  =  EF  (why?).     Q.  e.  d. 


Til.  XLII.]       THREE   OR  MORE  PARALLELS.  65 

84.  Def.  A  4-side  formed  by  two  parallels  and  two 
transversals  is  called  a  trapezoid.  Thus  ACFD  is  a  trape- 
zoid. The  parallel  sides  are  called  the  bases  (major  and 
min3r)  ;  the  parallel  through  the  mid-points  of  the  trans- 
verse sides  is  the  mid-parallel. 


Fig.  56. 

Theorem  XLII.  —  The  mid-parallel  of  a  trapezoid  equals 
the  half -sum  of  its  bases. 

Let  the  student  elicit  the  proof  from  the  foregoing  figure. 

Corollary  i .  A  parallel  to  a  base  of  a  A  bisecting  one 
side  bisects  also  the  other.     {Hint.     Let  D  fall  on  A.) 

Corollary  2.  A  ray  bisecting  two  sides  of  a  A  is  parallel 
to  the  third. 

For  only  one  ray  can  bisect  two  sides  (why?),  and  we 
have  just  seen  (Cor.  i)  that  a  ray  parallel  to  the  base  does 
this  ;  hence,     q.  e.  d. 

Corollary  3.  The  mid-parallel  to  the  base  of  a  A  equals 
half  the  base. 

85.  Def  Three  or  more  rays  that  pass  through  a  point 
are  said  to  concur  or  l)e  concurrent. 


66  GEOMETRY.  [Th.  XLIII. 

Theorem  XLIII.  —  The  me  dials  of  a  IS  concur. 

Data  :    ABC  a  A,  ^^aiid  BQ  two  medials  (Fig.  57). 


Proof.  Draw  a  ray  from  C  through  (9,  the  intersection 
of  the  two  medials,  and  lay  off  011=  CO.  Draw  AH  and 
BH;  they  are  parallel  io  BQ  and  AF  (why?)  ;  hence 
A  OBH  is  a  parallelogram  (why  ?)  ;  hence  AR  =  BR 
(why?).  Hence  COR  is  the  third  medial;  />.  the  three 
medials  pass  through  O.     q.  e.  d. 

Corollary.  Each  medial  cuts  off  a  third  from  each  of 
the  other  two.     For  C0  =  20R  (why?). 

Def.  The  point  of  concurrence  of  the  medials  is  called 
the  centroid  of  the  A.  It  is  two-thirds  the  length  of  each 
medial  from  the  corresponding  vertex. 

86.  Theorem  XLIV.  —  The  mid-normals  of  the  sides  of 
a  A  concur. 

Data :  ABC  a  A,  Z  and  M  mid-normals  to  the  sides  EC 
and  CA,  meeting  at  S. 


Th.  XLV.] 


CONCURRENTS, 


67 


Proof.  ^  is  equidistant  from  B  and  C  (why?),  and  from 
Cand  A  (why?)  ;  hence  S  is  equidistant  from  A  and  B 
(why?),  or  is  on  the  mid-normal  of  AB  (why?)  ;  hence 
the  mid-normals  concur  (Fig.  58).     g.  e.  d. 


Fig,  58. 


Corollary.  S  is  ecjuidistant  from  A,  B,  and  C,  and  no 
other  point  in  the  plane  is  (why?). 

D^/.  The  point  of  concurrence  of  the  mid-normals  is 
called  the  circumcentre  of  the  A. 

87.  Z>e/.  A  tract  from  a  vertex  of  a  A  normal  to  the 
opposite  side  is  called  an  altitude  of  the  A.  Sometimes, 
when  length  is  not  considered,  the  whole  ray  is  called  the 
altitude. 

Theorem  XLV.  —  T/ie  altitudes  of  a  ^  concur. 

Proof.  Using  the  preceding  figure,  draw  the  A  A'B'C. 
Its  sides  are  parallel  to  the  sides  of  ABC  (why?)  ;  hence 
its  altitudes  are  the  mid-normals  L,  Af,  JV;  and  these  have 
just  been  found  to  concur.  Also,  since  ABC  may  be  any 
A,  A'B'C  may  be  any  A  ;  hence  the  altitudes  of  any  A 
concur,     q.  e.  d. 


68  ■  GEOMETRY.  [Th.  XLVL 

Def.  The  point  of  concurrence  of  altitudes  is  called  the 
orthocentre  (or  alticentre)  of  the  A. 

Def.  In  a  right  A  the  side  opposite  the  right  angle  is 
called  the  hypotenuse  (=  subtense  =  under-stretch). 

Queries :  Where  do  circumcentre  and  orthocentre  lie : 
(i)  in  an  acute-angled  A?  (2)  in  an  obtuse-angled  A  ? 
(3)  in  a  right  A  ? 

88.  Theorem  XL VI. — The  inner  mid-rays  of  the  angles 
of  a  /S.  concur. 

Data :  ABC  a  A,  AL,  BM,  CiVthe  inner  mid-rays  of  its 
angles  (Fig.  59). 


Fig.  59. 

Proof.  Let  AL  and  BM  intersect  at  /.  Then  /  is  equi- 
distant from  AB  and  AC,  and  from  AB  and  BC  (why?)  ; 
hence  /  is  equidistant  from  AC  and  BC ;  hence  /  is  on 
the  inner  mid-ray  of  the  angle  C ;  i.e.  the  three  inner  mid- 
rays  concur  in  /.     q.  e.  d. 

Def  The  point  of  concurrence  of  the  inner  mid-rays  is 
called  the  in-centre  of  the  A. 


Th.  XLVII.]  EXERCISES  I.  69 

89.    Theorem  XLVII.  —  The  outer  mid-rays  of  hvo  angles 
and  the  inner  mid- ray  0/  the  other  angle  0/  a  A  concur. 
Let  the  student  conduct  the  proof  (Fig.  60). 


I 

Fig.  60. 

De/.  The  points  of  concurrence  are  called  ex-centres  of 
the  A  :  there  are  three. 

EXERCISES   I. 

Little  by  little  the  student  has  been  left  to  rely  more  and 
more  upon  his  own  resources  of  knowledge  and  ratiocination 
in  the  conduct  of  the  foregoing  investigations.  He  has  now 
possessed  himself  of  a  large  fund  of  concepts,  and  he  must 
test  his  ability  to  wield,  combine,  and  manipulate  them  in 
forging  original  proofs  of  theorems.  Let  him  bear  always 
in  mind  the  fundamental  logical  principle  that  ez'ery  example 


70  GEOMETRY. 

of  a  general  concept  has  all  the  marks  of  thai  general  concept. 
Let  him  begin  his  proof  by  stating  precisely  the  data,  the 
given  or  known  facts,  let  him  draw  a  corresponding  diagram 
in  order  to  have  a  clearer  view  of  the  spatial  relations  in- 
volved, let  him  note  carefully  what  concepts  are  present  in 
the  proposition,  let  him  draw  auxiliary  lines  and  introduce 
auxiliary  concepts  at  pleasure.  But  let  him  exhaust  simple 
means  before  trying  more  complicated,  let  him  distinguish, 
by  manner  of  drawing,  the  principal  from  the  auxiliary  rays, 
and  especially  let  him  be  systematic  and  consistent  in  the 
literation  of  his  figures. 

1.  How  many  degrees  in  a  straight  angle?  In  a  right 
angle  ? 

Historical  Note.  —  For  purposes  of  computation  the  round  angle 
is  divided  into  360  equal  parts  called  degrees,  each  degree  into  60 
equal  minutes  (partes  minutce  primse),  each  minute  into  60  equal 
seconds  (partes  minutae  secundcz),  denoted  by  *^,  ',  "  respectively. 
This  sexagesimal  division  is  cumbrous  and  unscientific,  but  is  apparently 
permanently  established.  It  seems  to  have  originated  with  the  Baby- 
lonians, who  fixed  approximately  the  length  of  the  year  at  360  days,  in 
which  time  the  sun  completed  his  circuit  of  the  heavens.  A  degree, 
then,  as  is  indicated  by  the  name,  which  means  step  in  Latin,  Greek, 
Hebrew  {gradus,  padfios  (or  rfXTjixa),  ma'a/ak),  was  primarily  the  daily 
step  of  the  sun  eastward  among  the  stars.  The  Chinese,  on  the  other 
hand,  determined  the  year  much  more  exactly  at  365}  days,  and 
accordingly,  in  defiance  of  all  arithmetic  sense,  divided  the  circle  into 
365 1  degrees. 

2.  The  angles  of  a  A  are  equal ;  how  many  degrees  in 
each? 

Remark.  —  Such  a  A  is  called  equiangular,  more  commonly  equi- 
lateral, but  better  still  regular. 

3.  Show  that  this  regular  A  is  equilateral. 

4.  One  angle  of  a  A  is  a  right-angle ;  the  others  are 
equal ;  how  many  degrees  in  each  ? 


EXERCISES  I.  71 

5.  One  angle  of  a  A  is  twice  and  the  other  thrice  the 
third  ;  what  are  the  angles  ? 

6.  Two  angles  of  a  A  are  measured  and  found  to  be 

46°  37' 24"  and  52° 48' 39";  what  is  the  third? 

7.  One  angle  of  a  A  is  measured  to  be  6i°22'4o";  the 
others  are  computed  to  be  49°  34'  28"  and  69°  2*43" ;  what 
do  you  infer  ? 

8.  A  half- ray  turns  through  two  round  angles  counter- 
clockwise, then  through  half  a  right-angle  clockwise,  then 
through  a  straight  angle  counter-clockwise,  then  through  \ 
of  a  round  angle  counter-clockwise,  then  through  |^  of  a 
straight  angle  clockwise  ;  what  angle  does  it  make  in  its 
final  position  with  its  original  position  ? 

9.  <9  is  a  fixed  point  (called  origin)  on  a  ray,  A  and  B 
are  any  pair  of  points,  M  their  mid-point.  Show  and  state 
in  words  that  2  0M=  OA  -f  OB. 

10.  Af  B^  C  are  three  points  on  a  ray,  A\  B\  C  are 
mid-points  of  the  tracts  BC^  CA,  AB,  and  O  is  any  point 
on  the  ray ;  show  that  OA-\-OB-\-  0C=  0A'-{-  0B'-\-  OC. 

11.  Af  By  C,  £>,  O  are  points  on  a  ray;  A',  B\  C  are 
mid-points  of  AB,  BC,  CD  \  A",  B",  are  mid-points  of 
A'B',  B'C;  Mis  the  mid-point  of  A"B" ;  prove  SOM= 
OA  +  sOB  +  sOC+OD. 

12.  What  are  the  conditions  of  congruence  in  isosceles 
A  ?     In  right  A  ? 

13.  In  what  A  does  one  angle  equal  the  sum  of  the 
other  two? 

Z>e/.  A  number  of  tracts  joining  consecutively  any  number 
of  points  (first  with  second,  second  with  third,  etc.)  is  called 
a  broken  line,  or  train  of  tracts,  or  polygon.     Where  the  last 


72  GEOMETRY. 

point  falls  on  the  first  the  polygon  is  closed ;  otherwise  it  is 
open.  Unless  otherwise  stated,  the  polygon  is  supposed  to 
be  closed.  The  points  are  the  vertices,  the  tracts  are  the 
sides  of  the  polygon.  The  closed  polygon  has  the  same 
number  of  vertices  and  sides,  and  we  may  call  it  an  n-angle 
or  n-side.  The  angles  between  the  pairs  of  consecutive 
sides  are  the  angles  of  the  polygon,  either  inner  or  outer ; 
unless  otherwise  stated,  inner  angles  are  referred  to.  Inner 
and  outer  angles  at  any  vertex  are  supplemental.  When 
each  inner  angle  is  less  than  a  straight  angle,  the  polygon 
is  called  convex  ;  otherwise,  re-entrant.  Unless  otherwise 
stated,  convex  polygons  are  meant.  Sides  and  angles  of  a 
polygon  may  be  reckoned  either  clockwise  or  counter-clock- 
wise. 

14.  Prove  that  the  sum  of  the  inner  angles  of  an  «-side 
is  {n  —  2)  straight  angles.  What  is  the  sum  of  the  outer 
angles  ? 

15.  Find  the  angle  in  a  regular  {i.e.  equiangular  and 
equilateral)  3-side,  4-side,  5-side,  8-side,  12-side.  (For 
proof  that  there  is  a  regular  ?z-side,  see  Art.  137.) 

16.  Show  that  a  (convex)  polygon  cannot  have  more 
than  three  obtuse  outer  angles,  nor  more  than  three  acute 
inner  angles. 

17.  Two  angles  of  a  A  are  a  and  ^\  find  the  angles  at 
the  intersection  of  their  mid-rays. 

18.  If  two  A  have  their  sides  parallel  or  perpendicular  in 
pairs,  then  the  A  are  mutually  equiangular. 

19.  The  medial  to  the  hypotenuse  of  a  right  A  cuts  the 
A  into  two  isosceles  A. 

20.  An  angle  in  a  A  is  obtuse,  right,  or  acute,  according 
as  the  medial  to  the  opposite  side  is  less  than,  equal  to,  or 
greater  than,  half  the  opposite  side. 


EXEKCISES  I.  n 

21.  A  medial  will  be  greater  than,  equal  to,  or  less  than, 
half  the  side  it  bisects,  according  as  the  opposite  angle  is 
acute,  right,  or  obtuse. 

22.  \{  P  and  Q  be  on  the  mid-normal  of  ABy  then 
AAPQ  =  ABPQ  (=  indicates  congruence). 

23.  AB  is  the  base,  C  the  opposite  vertex  of  an  isosce- 
les A;  show  that  ABN=BAM  (i)  when  AM  and  BN 
are  altitudes,  (2)  when  they  are  medials,  (3)  when  they  are 
mid-rays  of  angles  A  and  Z?,  (4)  when  MN  is  normal  to  the 
mid-normal  of  AB. 

24.  P  is  any  point  within  the  A  ABC;  show  that 
AP+BP<AC^  CB,  AP+PB-\-  CP>  ^ {AB+BC+  CA). 

25.  ABC"'L  and  AB'C-'-L  are  two  convex  polygons, 
not  crossing  each  other,  between  the  same  pairs  of  points, 
A  and  Z  ;  which  is  the  longer?     Give  proof. 

26.  P  is  a  point  within  AABC ;  show  that  angle  APB 
>ACB  and  sum  of  angles  at  P=  2{A  +  B-\-  C). 

27.  Pis  equidistant  from  A,  B,  and  C \  show  that  angle 
APB=2{zxi^^ACB). 

28.  Conversely,  if  angle  ^/^^  =  2  (angle  ACB),  angle 
BPC=  2  (angle  BAC),  and  angle  CPA  =  2  (angle  CBA), 
then  P  is  equidistant  from  A,  B,  C. 

29.  The  mid-rays  of  the  angles  at  the  ends  of  the  trans- 
verse axis  of  a  kite  cut  the  sides  in  the  vertices  of  an 
anti-parallelogram  (Art.  99). 

30.  The  four  joins  of  the  consecutive  mid-points  of  the 
sides  of  a  4-side  form  a  parallelogram. 

31.  The  joins  of  the  mid-points  of  the  pairs  of  opposite 
sides  and  of  the  pairs  of  diagonals  of  a  4-side  concur,  bisect- 
ing each  other. 


74  GEOMETRY. 

52.  The  mid-parallels  to  the  sides  of  a  A  cut  it  into  4 
congruent  A. 

33.  What  figures  are  formed  by  the  mid-parallels  when 
the  A  is  right?  isosceles?  regular? 

34.  A  parallelogram  is  a  rhombus  if  a  diagonal  bisects 
one  of  its  angles. 

35.  A  parallelogram  is  a  square  if  its  diagonals  are  equal 
and  one  bisects  an  angle  of  the  parallelogram. 

36.  From  any  point  in  the  base  of  an  isosceles  A  parallels 
are  drawn  to  the  sides ;  the  parallelogram  so  formed  has  a 
constant  perimeter  ( =  measure  round  =  sum  of  sides) . 

37.  The  sum  of  the  distances  of  any  point  on  the  base  of 
an  isosceles  A  from  the  sides  is  constant. 

38.  The  sum  of  the  distances  of  any  point  within  a  regular 
A  from  the  sides  is  constant. — What  if  the  point  be  without 
the  A? 

39.  /*  is  on  a  mid-ray  of  the  angle  A  in  the  A  ABC) 
compare  the  difference  of  FB  and  PC :  when  B  is  within 
the  A,  and  when  B  is  without. 

40.  The  inner  mid-ray  of  one  angle  of  a  A  and  the  outer 
mid-ray  of  another  form  an  angle  that  is  half  the  third  angle 
of  the  A. 

41.  O  is  the  orthocentre  of  the  A  ABC;  express  the 
angles  A  OB,  BOC,  CO  A,  through  the  angles  A,  B,  C. 

42.  Do  the  like  for  the  circum-centre  6" and  the  in-centre  /. 

43.  The  medial  to  the  hypotenuse  of  a  right  A  equals 
one-half  of  that  hypotenuse. 

44.  The  mid-rays  of  two  adjacent  angles  of  a  parallelogram 
are  normal  to  each  other. 


EXERCISES  I.  75 

45.  In  a  5 -pointed  star  the  sum  of  the  angles  at  the 
points  is  a  straight  angle.  What  is  the  sum  in  a  7-pointed 
star? 

46.  Parallels  are  drawn  to  the  sides  of  a  regular  A,  tri- 
secting the  sides ;  what  figures  result  ? 

47.  A  side  of  a  A  is  cut  into  8  equal  parts,  through  each 
section  point  parallels  are  drawn  to  the  other  sides  ;  how  are 
the  other  sides  cut  and  what  figures  result  ? 

48.  Two  A  are  congruent  when  they  have  two  mid-tracts 
of  two  corresponding  angles  equal,  and  besides  have  equal 

( 1 )  these  angles  and  a  pair  of  the  including  sides  ;  or 

(2)  two  pairs  of  corresponding  angles;  or 

(3)  one  pair  of  corresponding  angles  and  the  correspond- 
ing angles  of  the  mid-tract  with  the  opposite  side  ;  or 

(4)  one  pair  of  including  sides  and  the  adjacent  segment 
of  the  opposite  side. 

49.  Two  A  are  congruent  when  they  have  two  corre- 
sponding sides  and  their  medials  equal,  and  besides  have 
equal 

( 1 )  another  pair  of  sides ;  or 

(2)  the  angles  of  the  medial  with  its  side  (in  pairs) ;  or 

(3)  a  pair  of  angles  of  the  bisected  side  with  another 
side,  the  angles  of  the  medial  with  this  side  being  both 
acute  or  both  obtuse ;  or 

(4)  a  pair  of  angles  of  the  medial  with  an  including  side, 
the  corresponding  angles  of  the  medial  with  its  side  being 
both  acute  or  both  obtuse. 

50.  Two  A  are  congruent  when  they  have  a  pair  of  cor- 
responding altitudes  equal,  and  besides  have  equal 


76  GEOMETRY. 

( 1 )  the  pair  of  bases  and  a  pair  of  adjacent  angles  ;  or 

(2)  the  pair  of  bases  and  another  pair  of  sides  ;  or 

(3)  the  pairs  of  angles  of  the  altitude  with  the  sides ;  or 

(4)  two  pairs  of  corresponding  angles  ;  or 

(5)  the  two  pairs  of  sides,  when  the  altitudes  lie  both 
between  or  both  not  between  the  sides  of  the  A. 

SYMMETRY. 

90.  We  have  seen  that  congruent  figures  are  ahke  in  size 
and  shape,  different  only  in  place,  and  may  be  made  to  fit 
point  for  point,  line  for  line,  angle  for  angle.  The  parts 
that  fit  one  on  the  other  are  said  to  correspond  or  be  corre- 
spondent. Plainly  only  Hke  can  correspond  to  like,  as  point 
to  point,  etc. 

Def.  The  ray  through  two  points  we  may  call  the  join  of 
those  points,  and  the  point  on  two  rays  the  join  of  the  rays. 

91.  It  is  now  plain  that  if  A  corresponds  to  A  and  B  to 
B\  then  the  join  of  A  and  B  must  correspond  to  the  join 
of  A'  and  B^ ;  for  in  fitting  A  on  A'  and  B  on  B'  the  ray 
AB  must  fit  on  the  ray  A'B'  (why?).  Also  if  the  ray  Z 
corresponds  to  L',  and  Mto  M\  then  the  join  of  L  and  M 
must  correspond  to  the  join  of  V  and  J/'  (why?).  These 
facts  are  very  simple  but  very  important. 

We  shall  think  of  the  plane  as  a  thin  double  film,  the  one 
figure  drawn  in  the  upper  layer,  the  other  in  the  lower. 

92.  Two  congruent  figures  may  be  placed  anywhere  and 
any  way  in  the  plane,  but  there  are  two  positions  especially 
important:  (i)  the  one  in  which  the  one  figure  may  be 
superimposed  on  the  other  by  turning  the  one  half  of  the 
plane  through  a  straight  angle  about  a  ray  called  an  axis ; 
(2)  the  one  in  which  the  one  figure  may  be  fitted  on  the 


SYMMETRY. 


77 


Other  by  turning  the  one  half  of  the  plane  through  a  straight 
angle  about  a  j^oint  called  a  centre. 

Congnient  figures  in  either  of  these  two  positions  are 
called  symmetric  :  in  the  first  case  axally,  as  to  the  axis  of 
symmetry ;  in  the  second  case  centrally,  as  to  the  centre 
of  symmetry. 

93.  In  two  symmetries,  corresponding  angles,  like  all 
other  correspondents,  are  of  course  congruent ;  but  they 
are  reckoned  oppositely  if  the  symmetry  be  axal,  similarly  if 


Fig.  61. 


78  GEOMETRY. 

it  be  central.  To  parallels  correspond  parallels  ;  to  normals, 
normals  ;  to  mid-points,  mid-points  ;  to  mid-rays,  mid-rays  ; 
to  the  axis  corresponds  the  axis,  each  point  to  itself;  to  the 
centre  corresponds  the  centre  itself  (Fig.  6i). 

Elements,  whether  points  or  lines,  that  correspond  to 
themselves  may  be  called  self-correspondent  or  double. 

It  is  also  manifest  that  centre  and  axis  are  the  only  self- 
correspondents  ;  hence  if  a  point  be  self-correspondent,  it 
must  lie  on  the  axis  in  axal  symmetry,  or  be  the  centre  in 
central  symmetry ;  and  if  two  counter  half-rays  be  corre- 
spondent, they  (or  the  ray)  must  be  normal  to  the  axis  in 
axal  symmetry,  or  go  through  the  centre  in  central  sym- 
metry. 

94.  These  facts  are  all  perfectly  obvious,  but  a  more 
vivid  exemplification  of  the  nature  of  these  two  kinds  of 
symmetry  may  perhaps  be  found  in  the  following : 

Suppose  the  axis  of  symmetry  to  be  a  perfect  plane 
mirror ;  then  either  half  of  the  plane  may  be  treated  as  the 
reflection  or  exact  image  of  the  other,  and  will  be  the  sym- 
metric of  the  other  as  to  the  mirror-axis.  For  the  image  of 
any  point  A  is  the  point  A^  such  that  the  axis  is  the  mid- 
normal  of  AA\  as  we  know  froni  Physics ;  also,  on  folding 
over  the  one  half  of  the  plane  about  the  axis  upon  the  other 
half,  the  point  A  falls  on  A^  (why?)  ;  hence  A^  is  the  sym- 
metric of  A  as  to  the  axis. 

Suppose  the  centre  of  symmetry  S  to  be  also  a  reflector ; 
then  the  reflection  or  image  of  any  point  A  will  be  a  point 
A^  such  that  6"  is  the  mid-point  of  the  tract  AA\  and  on 
rotation  through  a  straight  angle  about  6"  the  point  A  falls 
on  A\  and  the  half-ray  SA  fits  on  the  half- ray  SA\  Hence 
either  of  two  centrally  symmetric  figures  is  the  exact  image 
of  the  other  reflected  from  the  centre  of  symmetry  6". 


SYMMETRY. 


79 


Note  carefully  that  these  two  species  of  symmetry  depend 
upon  the  two  fundamental  definitive  properties  of  the  plane  : 
central  symmetry  upon  the  homa'oidality  of  the  plane,  axal 
symmetry  upon  the  reversibility  of  the  plane.  Moreover, 
axally  symmetric  figures  can  7iot  be  fitted  on  each  other 
without  reversion,  folding  over ;  by  movement  in  the  plane 
their  corresponding  parts  can  at  best  be  ^/posed,  but  never 
j«/^rposed ;  while  on  the  other  hand  central  symmetries 
may  be  j/y/^rposed,  but  cannot  be  <7/posed,  along  any  ray, 
by  motion  in  the  plane.  In  central  symmetries  the  corre- 
sponding parts  follow  one  another  in  the  same  order,  but  in 
axal  symmetries  they  follow  in  opposite  orders. 

95.  We  must  now  discuss  these  two  symmetries  more 
minutely,  and  to  exhibit  a  certain  remarkable  relation  hold- 
ing between  them  we  arrange  their  properties  in  parallel 
columns. 


In  Axal  Symmetry. 

1.  The  axis  corresponds  to  it- 
self. 

2.  Every  point  of  the  axis  cor- 
responds to  itself. 

3.  Every  self-correspondent 
point  lies  on  the  axis. 

4.  The  join  of  two  correspond- 
ent rays  is  on  the  axis. 

(For  it  is  self-correspondent.) 

5.  Correspondent  points  are 
equidistant  from  every  point  on 
the  axis. 


In  Central  Symmetry. 

1.  The  centre  corresponds  to 
itself. 

2.  Every  ray  through  the  cen- 
tre corresponds  to  itself  (each  half 
to  the  other). 

3.  Every  self-correspondent  ray 
goes  through  the  centre. 

4.  The  join  of  two  correspond- 
ent points  goes  through  the  cen- 
tre. 

(For  it  is  self-correspondent.) 

5.  Correspondent  rays  are 
equally  inclined  (isoclinal)  to 
every  ray  through  the  centre; 
hence  they  are  parallel,  as  is 
otherwise  manifest. 


80 


GEOMETRY. 


6.  The  centre  is  a  mid-point  of 
every  tract  between  correspond- 
ent points,  and  in  fact  the  inner 
mid-point. 

N.B.  The  outer  mid-point  is  a 
point  at  infinity. 

7.  The  join  of  two  correspond- 
ent rays  is  at  infinity. 

(For  they  are  parallel.) 

8.  Correspondent  angles  are 
contra-posed  {i.e.  have  their  arms 
extended  oppositely). 

9.  Correspondent  rays  are 
equidistant  from  the  centre. 

10.  The  join  of  two  points  and 
the  join  of  their  correspondents 
themselves  correspond. 


6.  The  axis  is  a  mid-ray  of 
every  angle  between  correspond- 
ent rays,  and  in  fact  the  inner 
mid-ray. 

N.B.  The  outer  mid-ray  is  a 
normal  to  the  axis. 

7.  The  join  of  two  correspond- 
ent/o2«/^  is  a  normal  to  the  axis. 

8.  Correspondent  tracts  are 
anti-parallel. 

9.  Correspondent  points  are 
equidistant  from  the  axis. 

10.  The  join  of  two  rays  and 
the  join  of  their  correspondents 
themselves  correspond. 

96.  On  regarding  closely  these  correlated  propositions,  it 
becomes  clear  that  the  one  set  differs  from  the  other  only 
in  the  interchange  of  certain  notions,  as  poi7it  and  ray,  tract 
and  angle,  etc.  Every  property  of  axal  symmetry  has  its 
obverse  in  central  symmetry,  and  vice  versa.  This  most 
profound,  important,  and  interesting  fact  has  received  the 
name  of  the  Principle  of  Reciprocity.  We  make  this  notion 
more  precise  by  the  following 

Def.  Two  figures  such  that  to  every  point  of  each  corre- 
sponds a  ray  of  the  other,  and  to  every  ray  of  each  a  point 
of  the  other,  are  called  reciprocal.     For  example  : 

Suppose  rays  drawn  through  a  point  O  to  any  number  of 
points.  A,  B,  C,  D,  E,  .  .  .  on  2i  ray  L.  Then  the  point 
O  with  its  ray  through  it,  and  the  ray  L  with  its  point  on 
it,  are  two  reciprocal  figures  (Fig.  62).  The  first  is  called 
a  (flat)  pencil  of  rays,  O  being  the  centre ;  the  second  is 
called  a  row  (or  range)  of  points,  L  being  the  axis.     Sup- 


SYMMETRY. 


81 


pose  we  have  now  a  second  pencil  through  O'  and  a  second 
row  on  Z'.  These  two  figures  are  again  reciprocal,  and  the 
two  pairs  of  reciprocals  together  make  up  another  more 
complex  pair  of  reciprocals.  In  this  latter  pair  we  find  our 
definition  fully  exemplified.  To  O  and  6>'  correspond  L  and 
Z' ;  to  the  rays  through  O  and  (9'  correspond  the  points  on  Z 
and  Z' ;  also,  to  the  join  (ray)  of  O  and  O'  corresponds  the 


Fig.  62. 

join  (point)  of  Z  and  Z' ;  to  any  point  as  Pj  the  join  of  two 
rays  {OA^  O'A'),  corresponds  a  ray  A  A',  the  join  of  two 
points  {A J  A').  So  Q,  R^  S,  Tare  points  corresponding  to 
the  rays  BB\  CC\  DD\  EE\  We  may  notice  further 
that  angle  and  tract  correspond  in  the  reciprocal  figures ; 
thus  the  angle  AOB  corresponds  to  the  tract  AB,  and  the 
angle  BOC  to  the  tract  BC\  while  the  angle  OPO'  corre- 
sponds to  the  tract  AA^  and  the  tract  RS  to  the  angle 
between  the  rays  corresponding  to  R  and  S ;  namely,  between 
CO  and  DD\  Let  the  student  trace  out  as  many  corre- 
spondences as  possible. 


82  GEOMETRY, 

97.  To  three  points  fixing  a  triangle  in  either  of  two 
reciprocals  must  correspond  also  three  rays  fixing  a  triangle 
in  the  other  reciprocal;  hence,  in  general,  triangle  corre- 
sponds to  triangle  in  reciprocals.  But  notice  :  the  sides  of 
one  correspond  to  the  vertices  of  the  other ;  hence  if  the 
sides  of  one  all  go  through  the  same  point,  the  vertices  of 
the  other  all  lie  on  the  same  ray ;  that  is,  three  concurrent 
rays  in  either  reciprocal  correspond  to  three  collinear  points 
in  the  other. 

It  now  appears  that  axal  and  central  symmetry  are  recip- 
rocal to  each  other ;  the  reciprocal  of  an  axal  symmetric  is 
a  central  symmetric,  and  the  reciprocal  of  a  central  sym- 
metric is  an  axal  symmetric ;  the  reciprocal  properties  of 
axal  symmetry  are  the  properties  of  central  symmetry,  and 
the  reciprocal  properties  of  central  symmetry  are  the  proper- 
ties of  axal  symmetry. 

Very  often  the  two  symmetric  figures  may  be  regarded 
as  the  two  halves  of  one  figure  ;  this  one  figure  is  then  said 
to  be  symmetric  as  to  the  axis  of  symmetry  or  as  to  the 
centre  of  symmetry,  as  the  case  may  be. 

98.  If  our  figure  be  two  points,  A  and  A\  then  the  mid- 
normal  X  of  the  tract  A  A  is  the  axis  of  symmetry,  mani- 
festly. If,  now,  any  double  point  D  on  the  axis  be  joined 
with  A  and  A\  there  results  the  isosceles  A  ADA\  whence 
it  appears  that  (Fig.  ^2>) 

The  isosceles  A  is  a  symj?tetric  A. 

It  is  plain  that  any  two  points  on  the  ray  AA^  equidistant 
from  N  are  symmetric  as  to  X,  that  all  points  on  the  ray, 
and  indeed  in  the  whole  plane,  may  be  arranged  in  sym- 
metric pairs,  the  members  of  each  pair  equidistant  from  the 
axis  X. 


SYMMETRY. 


83 


99.  Now  take  two  points  on  the  axis,  as  D  and  D\  or 
D  and  Z>",  and  consider  the  4-side  DAD'A^  It  is  com- 
posed of  two  A,  ADiy  and  ADD\  symmetric  with  each 
other  as  to  the  axis  X,  and  opposed  along  that  axis.  Hence 
the  4-side  is  itself  symmetrical  as  to  X. 

Def.  Such  a  4-side,  with  an  axis  of  symmetry,  is  called  a 
kite. 

If  we  hold  D  fast,  and  let  /)'  glide  along  Jf,  the  4-side 
ADA'iy  remains  a  kite.     We  see  that  there  are  two  kinds 


of  kites,  the  convex  kite,  as  ADA'Z>\  and  the  re-entrant,  as 
ADA'D".  As  the  gliding  point  passes  through  TV  the  kite 
changes  from  one  kind  to  the  other,  passing  through  the 
intermediate  form  of  the  symmetrical  A. 

When  the  gliding  point  reaches  a  position  £>'  such  that 
JVD  =  JVD',  then  the  four  sides  of  the  kite  are  all  equal 
(why?),  and  the  kite  becomes  a  rhombus  (why?).  In  this 
case  £>  and  D'  are  symmetric  as  to  A  A'  as  an  axis  of  syni- 


84 


GEOMETRY. 


metry.      Hence  the  rhombus   has  two  axes  of  syf?itnetry ; 
namely,  its  two  diagonals. 

In  all  cases  the  diagonals,  A  A'  and  DD\  of  the  kite  are 
normal  to  each  other  (why?). 

100.    Now  consider  a  pair  of  points,  B  and  B\  symmetric 
as  to  the  axis  X  (Fig.  64).     Then  X  is  mid-normal  oi  BB'. 


C 


Fig.  64, 

If  C  and  C  be  any  other  pair  of  symmetric  points,  then  X 
is  also  mid-normal  of  CC  ;  hence  BB^  and  CC  are  parallel 
(why?).  Also  the  tracts  BC  and  B' C  are  symmetric  as 
to  X  (why?),  and  the  4-side  BB'CC  is  itself  symmetric  as 
to  the  axis  X.     Hence  the  angles  at  C  and  C  are  equal, 


SYMMETRY. 


85 


also  the  angles  at  B  and  B'  are  equal  (why  ?)  ;  hence  the 
angles  at  B  and  C  and  at  B'  and  C  are  supplemental 
(why?),  and  the  4-side  BB'CC  is  an  anti-parallelogram 
(why?).  Hence  we  see  that  another  sy  nunc  trie  ^-side  is  an 
anti-parallelogram. 

It  is  plain  that  every  anti- parallelogram  is  symmetric,  for 
we  know  that  the  oblique  sides  prolonged  yield  an  isosceles  A. 
Let  the  student  complete  the  proof. 

loi .  There  is  only  one  kind  of  symmetric  A,  the  isosceles. 
For,  let  ABA'  (Fig.  65)  be  symmetric  and  A'  correspondent 


Fig.  65. 

to  A.  Then  B  must  correspond  to  itself  (why?)  ;  hence 
B  must  lie  on  the  axis  (why?)  ;  hence  BA  =  BA'  (why?). 

Now  let  the  student  prove  that 

(i)  /n  a  symtnetric  A  the  axis  of  symmetfy  is  a  medial ; 
(2)   //  is  also  a  mid-ray ;   (3)   /'/  is  also  a  mid-normal. 


86  GEOMETRY. 

Conversely,  let  him  show  that 

A  medial  that  is  a  mid-ray,  or  a  mid-normal,  is  an  axis 
of  symmetry. 

102.  There  are  only  two  axally  symmetric  4-sides  ;  namely, 
the  kite  and  the  anti- parallelogram.  For,  in  a  symmetric 
4-side  a  vertex  must  correspond  to  a  vertex  (why?).  Also, 
not  all  vertices  can  be  on  the  axis  (why?).  Also,  a  vertex 
on  the  axis  is  a  double  point  (why?).  Also,  the  vertices 
not  on  the  axis  must  appear  in  pairs  (why  ?)  ;  hence  there 
must  be  either  two  or  four  of  them.  If  there  be  two  only, 
then  the  other  two  are  on  the  axis  and  the  4-side  is  a  kite ; 
if  there  be  four  of  them,  we  have  just  seen  that  the  4-side  is 
an  anti-parallelogram. 

103.  Now  let  us  turn  to  the  reciprocals.  The  reciprocals 
of  the  two  points  A  and  A^  symmetric  as  to  the  axis  X  will 
be  two  rays  L,  L\  symmetric  as  to  the  centre  S.  But  rays 
symmetric  as  to  a  centre  are  parallel  (why  ?)  ;  hence  we  have 
two  parallels  symmetric  as  to  S,  which  is  midway  between 
them.  The  rays  are  symmetric  as  to  any  other  point  S'  mid- 
way between  them  (why?).  The  piece  of  plane  between 
these  parallels  is  called  a  parallel  strip,  or  band  (Fig.  66). 


R,' 


y^ 

L 

•s' 

L 

Fig.  66. 


SYMMETRY.  87 

But  what  corresponds  to  the  point  D  on  the  axis  X?  The 
answer  is  :  a  ray  R  through  6"  (why?).  Hence  to  the  sym- 
metric A  of  the  three  points  A,  A\  D,  there  corresponds 
the  figure  formed  by  two  parallels  Z,  Z',  and  a  transverse  R 
through  Sf  —  a  so-called  half-strip.  This  is  truly  a  three- 
side,  but  not  apparently  a  A  (3-angle),  for  the  parallels  do 
not  meet  in  finity,  in  regions  accessible  to  our  experience. 
Hence,  instead  of  saying  that  the  reciprocal  of  a  A  in  axal 
symmetry  is  a  A  (3-angle  or  3-point)  in  central  sym- 
metry, we  should  have  said,  accurately,  that  the  recipro- 
cal of  a  A  in  axal  symmetry  is  a  3-side  (or  trilateral)  in 
central  symmetry,  which  will  always  be  a  A  except  when 
sides  are  parallel  or  all  concur.  In  higher  Geometry  it  is 
very  convenient  to  remove  this  apparent  exception  by  using 
this  form  of  expression  :  the  parallels  meet  not  in  finity y  but 
///  infinity. 

104.  It  is  indeed  plain  that 

A  A  can  have  no  centre  of  symmetry. 

For,  since  vertex  corresponds  to  vertex,  and  since  corre- 
spondents appear  in  pairs,  one  vertex  must  be  a  double  point ; 
hence  it  would  have  to  be  the  centre  .S"  (why?).  But  the 
other  two  vertices  would  have  to  lie  on  a  ray  through  6",  being 
correspondents  ;  hence  the  three  vertices  would  be  collinear, 
and  the  A  would  be  flattened  out  to  a  triply-laid  ray. 

105.  But  there  is  a  centrally  symmetrical  4-side  ;  namely, 
the  parallelogram.  For,  consider  once  more  the  kite 
AXA'X'  and  let  us  reciprocate  it  into  a  centrally  symmetric 
figure  (Fig.  67).  To  the  axis  XX'  will  correspond  the  cen- 
tre S;  to  the  symmetric  pair  of  rays  AX  and  A'X  w'\\\  corre- 
spond a  symmetric  pair  of  points  Pand  P  \  to  the  join  of 
those  on  the  axis  A' will  correspond  the  join  of  these  through 


88 


GEOMETRY. 


the  centre  {PP^).  Similarly,  to  the  symmetric  rays  AX'  diud 
A'X'  will  correspond  the  symmetric  points  Q  and  Q',  and 
to  the  join  X'  will  correspond  the  join  QQ'.  Also,  AX 
and  AX'  have  a  join  A  while  A'X  and  A'X'  have  a  join  A', 
and  these  joins  are  symmetric  as  to  the  axis  XX' ;  recipro- 


FlG.  67. 

cally,  Pand  Q  have  a  join  PQ,  and  /"  and  Q'  have  a  join 
/"^'j  and  these  joins  are  symmetric  as  to  .S";  that  is,  they  are 
parallel  (why?).  Similarly,  PQ'  and  P'Q  correspond  to  B 
{AX,  A'X')  and  B'  {A'X,  AX')  ;  but  B  and  B'  are  sym- 
metric as  to  JTX'  (why?)  ;  hence  PQ'  and  P'Q  a.re  symmetric 
as  to  S,  i.e.  are  parallel.  Hence  PQ  PQ  is  symmetric  as 
to  S,  and  is  2,  parallelogram.     Q.  e.  d. 


106.    We  may  indeed  see  at  once  that  since  any  two  par- 
allels are  centrally  symmetrical  as  to  any  mid-point,  a  pair 


SYMMETRY, 


89 


of  parallels  or  a  parallelogram  is  symmetric  as  to  the  common 
mid-way  point,  the  intersection  of  the  diagonals.  But  the 
foregoing  reciprocation  is  instructive,  as  illustrating  in  detail 
the  method  to  be  pursued,  and  as  showing  the  intimate  rela- 
tion of  the  different  symmetric  quadrilaterals ;  namely,  the 
parallelogram  is  the  common  reciprocal  of  both  kite  <///// anti- 
parallelogram,  which  are  thus  seen  to  be  really  one, 

107.  Central  symmetry  does  not  in  general  imply  any- 
thing at  all  with  respect  to  axal  symmetry  in  a  figure.  We 
may  draw  through  any  point  S  any  number  of  rays  and  lay 
off  on  each  from  5  a  pair  of  counter  tracts  SP  and  SP\ 
SQ  and  SQ^  etc.  No  matter  how  PQ^  etc.,  be  chosen,  the 
figure  so  obtained  will  be  centrally  symmetric  as  to  6";  but 
it  may  have  no  axal  symmetry  whatever.  Neither  does  axal 
symmetry  in  general  imply  any  central  symmetry,  but  we 
may  establish  the  following  important 

Theorem.  —  Any  figure  with  two  rectangular  axes  of 
symmetry  has  also  a  centre  of  symmetry;  namely^  the  inter- 
section of  those  axes. 


p 

^ 

f 

p 

'^^ 

S/-^ 

X 

^^ 

1 

w- 

p 

Y 

Fiu.  68. 


90  GEOMETRY. 

Data :  XX^  and  KF'  two  rectangular  axes,  P  any  point 
of  a  figure  symmetric  as  to  these  axes  (Fig.  68). 

Proof.  The  point  /"  symmetric  with  P  as  to  XX  is  a 
point  of  the  figure  (why?)  ;  also  /*"  symmetric  with  /"  as 
to  yy  is  a  point  of  the  figure  (why?)  ;  so  too  is  /*'" 
(why?)  ;  the  figure  /'/"/'"P"'  is  a  rectangle  (why?),  its 
diagonals  halve  each  other,  and  SP=  SP"  =  SP' =  SP'". 
Hence  6*  is  a  centre  of  symmetry,     q.  e.  d. 

THE   CIRCLE. 

io8.  We  have  already  discovered  the  existence  of  a 
homceoidal  plane  curve  not  reversible  and  have  named  it 
circle. 

Defs.  A  ray  cutting  a  curve  is  called  a  secant,  as  L ;  the 
part  of  the  secant  intercepted  by  the  curve,  or  the  tract 
between  two  points  of  the  curve,  is  called  a  chord,  as  AB. 
A  finite  part  of  a  curve  is  called  an  arc.  A  chord  and  an 
arc  with  the  same  two  ends  are  said  to  subtend  each  other. 
Also,  the  intercept  of  any  line  between  the  ends  of  an  angle 
is  said  to  subtend  \h.&  angle.  Thus  .^Cand  DE  subtend  the 
angle  O  (Fig.  69). 


Fig.  69. 


Tn.  XLIX.J 


THE   CIRCLE. 


91 


109.  Theorem  XLVIII.  —  Congruent  arcs  subtend  con- 
gruent chords. 

Proof.  Let  the  arcs  AB  and  CD  be  congruent ;  then  we 
may  fii  A  on  C  and  at  the  same  time  B  on  £>;  then  the 
chords  AB  and  CB  fit  throughout  (why?),     q.  e.  d. 

N.B.  We  can  no/  convert  this  proposition  at  once  (why?) 
(Fig.  70). 


Fig.  70. 

1 10.    Theorem  XLIX.  — A  closed  curve  is  cut  by  a  ray  in 
an  eiien  number  of  points  (Fig.  71). 


Proof.  Let  Z  be  a  ray,  C  any  closed  curve.  Suppose  a 
point  P  to  trace  out  the  ray  Z.  At  first  P  is  without  the 
curve,  at  last  it  is  also  without  the  curve ;  hence  P  has 
crossed  the  curve  going  out  as  often  as  it  has  crossed  the 
curve  going  in,  for  every  entrance  there  is  an  exit ;  hence 
the  points  of  intersection  appear  in  pairs,  their  number  is 
even,  as  o,  2,  4,  6,  ...  2  //.     q.  e.  d. 


92 


GEOMETRY. 


[Til   L. 


These  preliminary  or  auxiliary  theorems,  which  prepare 
the  way  for  a  theorem  to  follow,  are  sometimes  called 
lemmas  {Xrjfxfxa  =  assumption,  premise,  support,  prop). 

*iii.  Theorem  L. — A  chrle  has  aft  axis  of  symmetry 
through  every  one  of  its  points  (Fig.   72). 


Q 


D 
R 


Q 


Fig.  72. 

Proof.  Let  D  be  any  point  of  a  circle.  Take  any  arc 
DP,  and  slip  it  round  till  P  falls  on  D  and  D  on  /" ;  this 
is  possible  (why  ?) .  Then  PDP'  is  a  symmetrical  A  (why  ?)  ; 
and  its  axis  of  symmetry  DR  halves  normally  the  chord 
PP',  and  also  halves  the  angle  PDP'  (why?).  Now  take 
any  other  arc  DQ  and  slip  it  round  till  Q  falls  on  D  and  D 
on  Q\  so  that  DQ  and  QD  are  congruent.  Then  the 
chords  DQ  and  DQ  are  congruent  (why?).  Also,  on 
taking  away  the  congruents  DP  and  DP'  we  have  left  PQ 
and  P'Q  as  congruent  remainders.     Hence  the  chords  PQ 


Th.  LI.]  THE    CIRCLE.  93 

and  PQ  are  congruent  (why?).  Hence  the  A  PDQ  and 
PDQ'  are  congruent  (why?)  ;  hence  the  angles  PDQ  and 
P'DQ'  are  equal  (why?)  ;  hence  DR  halves  also  the  angle 
QDQ  (why?).  But  the  A  QDQ  is  symmetric  (why?)  ; 
hence  DR  is  also  its  axis  of  symmetry,  and  Q  and  Q  are 
symmetric  points  of  the  circle ;  hence  any  point  of  the 
circle  has  its  symmetric  point  as  to  DR ;  i.e.  DR  is  an 
axis  of  symmetry  of  the  circle.  Moreover,  D  was  any  point 
of  the  circle  ;  hence  through  any  point  of  the  circle  passes 
an  axis  of  symmetry,     q.  e.  d. 

Def.  A  ray  halving  a  system  of  parallel  chords  is  called  a 
diameter ;  the  chords  and  diameter  are  called  conjugate  to 
each  other. 

Corollary  i.  In  a  circle  a  diameter  is  normal  to  its  con- 
jugate chords. 

Corollary  2.  Every  mid-normal  to  a  chord  in  a  circle  is 
a  diameter  and  halves  the  subtended  arcs. 

*ii2.  Theorem  LI.  —  A  circle  has  a  centre  of  symmetry 
(Fig.  12). 

For  the  ray  through  D  must  cut  the  circle  in  some  second 
point,  as  R  (why?),  and  as  the  ray  turns  round  from  the 
position  DR  to  the  reversed  position  RD,  through  a  straight 
angle,  it  must  pass  through  some  position,  QQy  normal  to 
its  original  position  (why  ?) .  Hence  for  any  axis  of  symmetry 
there  is  another  normal  thereto  and  their  intersection  is  a 
centre  of  symmetry  (why  ?) .    q.  e.  i>. 

N.B.   There  is  only  one  centre  of  symmetry  (why?). 

Def.  This  centre  of  symmetry  is  named  centre  of  the 
circle.  It  is  often  convenient  to  call  the  whole  ray  through 
the  centre  a  centre  ray  or  line,  and  to  restrict  the  term 
diameter  to  the  centre  chord. 


94 


GEOMETRY. 


[Th.  LII. 


Corollary  i.  All  diameters  go  through  the  centre,  and 
halve  each  other  there ;  conversely,  chords  halving  each 
other  are  diameters. 

Def.  Two  diameters  each  halving  all  the  chords  parallel 
to  the  other  are  called  conjugate. 

Corollary  2.  In  the  circle  two  diameters  normal  to  each 
other  are  conjugate  ;  and  conversely,  two  conjugate  diameters 
are  normal  to  each  other. 

N.B.  Other  curves,  as  ElHpse  and  Hyperbola,  have 
conjugate  diameters  not  in  general  normal  to  each  other 
(Fig-  73)- 

*ii3.  Theorem  LII.  —  All  diameters  of  a  circle  are  equal 
(Fig.  74). 


Fig.  73.  Fig.  74. 

Proof.  Let  DR  and  D^R^  be  two  diameters.  The  figure 
DD^RR^  is  a  parallelogram  (why?),  and  DD^  is  parallel  to 
RR^ ;    hence  the  mid-normal  of  these  parallels  is  a  diameter 


Th.  LI  I.]  THE    CIRCLE.  95 

through  the  centre  ^ ;    hence  SD  and  SD^  are  symmetric 
and  equal ;  hence  DR  =  D'R'.     q.  e.  d. 

De/.  A  half- diameter,  from  centre  to  circle,  is  called  a 
radius. 

Corollary  i.  All  radii  of  a  circle  are  equal ;  or,  all  points 
of  a  circle  are  equidistant  from  the  centre. 

Corollary  2.  Every  parallelogram  inscribed  in  a  circle  is 
a  rectangle. 

N.B.  By  help  of  this  important  property  the  circle  is 
commonly  defined  as  a  plane  curiae  all  points  of  which  arc 
equidistant  from  a  point  within  called  the  centre.  The  com- 
mon distance  of  all  points  of  the  circle  from  the  centre  is 
often  called  the  radius.  We  have  deduced  this  property 
from  the  homoeoidality ;  conversely ^  we  may  deduce  the 
homoeoidality  from  this  property  taken  as  definition.  But 
if  there  were  no  such  surface  as  the  plane,  at  least  for  our 
intuition,  the  circle  might  still  exist  on  the  sphere-surface, 
without  centre,  but  with  the  body  of  its  properties  unimpaired. 
Hence  it  seems  better  to  define  the  circle  by  its  intrinsic 
homoeoidality  than  by  its  extrinsic  centrality. 

Corollary  i.  All  points  within  the  circle  are  less,  and  all 
points  without  are  more,  than  the  radius  distant  from  the 
centre. 

Defs.  The  two  symmetric  halves  into  which  a  diameter 
cuts  a  circle  are  called  semicircles.  The  part  of  the  plane 
bounded  by  an  arc  and  its  chord  is  called  a  segment ;  the 
part  bounded  by  an  arc  and  the  two  radii  to  its  ends  is 
called  a  sector.  If  the  sum  of  two  arcs  be  a  circle,  we  may 
call  them  explemental,  the  one  minor y  the  other  major; 
every  chord  belongs  equally  to  each  of  two  explemental  arcs, 
but  in  general,  unless  otherwise  stated,  it  is  the  minor  that 


96  GEOMETRY.  [Th.   LIII. 

is  referred  to.  Two  arcs  whose  sum  is  a  half-circle  are 
called  supplemental ;  two  whose  sum  is  a  quarter-circle  or 
quadrant  are  called  complemental. 

Corollary  2.  All  circles  of  the  same  radius  are  congruent ; 
also,  all  semicircles  of  the  same  radius  are  congruent,  and 
all  quadrants  of  the  same  radius  are  congruent. 

Corollary  3.  Any  circle  may  be  shpped  round  at  will 
upon  itself  about  its  centre  as  a  pivot,  like  a  wheel  about  its 
axle,  without  changing  in  the  least  the  position  of  the  whole 
circle. 

114.  From  the  foregoing  it  is  clear  that  if  we  hold  one 
point  of  a  ray  fixed,  and  turn  the  ray  in  the  plane  about  the 
fixed  point,  every  other  point  of  it  will  trace  out  a  circle 
about  the  fixed  point  as  a  centre.  An  instrument,  one  point 
of  which  may  be  fixed  while  the  other  is  movable  about  in  a 
plane,  is  called  a  compass  or  pair  of  compasses,  and  is  both 
the  simplest  and  the  most  important  of  all  instruments  for 
drawing. 

115.  Theorem  LIII.  —  Through  any  three  points  not  col- 
linear  one,  and  only  one,  circle  f?iay  be  drawn. 

Proof.  Let  A,  B,  C  be  the  three  points  not  collinear 
(Fig.  75).  We  have  already  seen  that  the  mid-normals  to 
the  tracts  AB,  EC,  CA  concur  in  a  point  S  equidistant  from 
A,  B,  and  C ;  hence  a  circle  about  S  with  radius  d  passes 
through  A,  B,  C.  Also  there  is  only  one  point  thus  equi- 
distant from  A,  B,  C  (why?)  ;  hence  there  is  only  one 
circle  through  A,  B,  C.     Q.  e.  d. 

Def.  The  circle  through  the  vertices  A,  B,  C,  of  a  A  is 
called  the  circum-circle  of  the  A. 

Corollary  i.  A  A,  or  a  triplet  of  points,  or  a  triplet  of 
rays,  determines  one,  and  only  one,  circle. 


Th.   Mil.] 


THE   CIRCLE. 


97 


Corollary  2.  Through  two  points,  A  and  B,  any  number 
of  circles  may  be  drawn.  Their  centres  all  lie  on  the  mid- 
normal  of  AB. 

Corollary  3.  ks  BC  turns  clockwise  about  i9  as  a  pivot, 
the  intersection  S,  the  centre  of  the  circle  through  A^  B,  C, 
retires  upward  ever  faster  and  faster  along  the  mid-normal  N 
of  AB ;  when  C  becomes  collinear  with  A  and  B,  the  inter- 


FIG.  75. 

section  of  the  raid-normals  of  AB  and  BC  vanishes  from 
finity,  or  rf/ires  to  infinity,  as  the  phrase  is.  As  BC  keeps 
on  turning,  S  reappears  in  finity  below  and  moves  slower  and 
slower  upward  along  the  mid-normal.  Moreover,  a  circle 
passes  through  A,  By  and  C,  no  matter  how  close  C  may  lie 
to  the  ray  AB,  nor  on  which  side  of  it :  only  as  C  falls  upon 
the  ray  does  the  centre  of  the  circle  vanish  into  infinity ;  that 
is,  we  may  draw  a  circle  that  shall  fit  as  close  to  the  ray  AB 
as  we  please f  though  not  upon  it,  by  retiring  the  centre  far 


98  GEOMETRY.  [Th.  LTV. 

enough.  Hence  a  ray  may  be  conceived  as  a  circle  with 
centre  retired  to  infinity ;  it  is  strictly  the  limit  of  a  circle 
whose  centre  has  retired,  along  a  normal  to  it,  without  limit. 

1 1 6.  Theorem  LIV.  — A  circle  can  cut  a  ray  in  only  two 
points. 

For  there  are  only  two  points  on  a  ray  at  a  given  distance 
from  a  fixed  point  (why  ?) .     Q.  e.  d. 

117.  Theorem  LV.  —  Secants  that  make  equal  angles  with 
the  centre  ray  (or  axis)  through  their  intersection  intercept 
equal  arcs  on  the  circle. 

Proof.  For  both  the  two  semicircles  and  the  two  secants 
are  symmetric  as  to  the  axis  IS  (why  ?)  ;  hence,  on  folding 
over  the  one  half-plane  upon  the  other,  A  falls  on  A\  B  on 
B\  arc  a  fits  on  arc  a\  and  chord  c  on  chord  c^  (Fig.  76). 

Q.  E.  D. 


Tn.  LVI.]  THE   CIRCLE.  99 

Conversely,  Secants  thai  intercept  equal  arcs  make  equal 
angles  with  the  axis  through  their  intersection. 

Proof.  Let  L  and  V  intersect  equal  arcs  AB  and  A^B\ 
Draw  the  mid-normal  of  AA^ ;  it  is  an  axis  of  symmetry 
(why?).  On  folding  over  the  left  half- plane  upon  the  right 
half-plane,  A  falls  on  A  and  B  onB'  (why?)  ;  hence  AB 
and  A'B'  are  symmetric ;  hence  they  meet  on  the  axis  and 
make  equal  angles  with  it  (why?),     q.  e.  d. 

Corollary  i.  Equal  chords  are  equidistant  from  the  cen- 
tre ;  and  conversely ^  Chords  equidistant  from  the  centre  are 
equal. 

Corollary  2.  The  greater  of  two  unequal  chords  is  less 
distant  from  the  centre. 

Corollary  3.   A  diameter  is  the  greatest  chord. 

Corollary  4.  Arcs  intercepted  by  two  parallel  chords  are 
equal. 

Corollary  5.  Equal  chords  or  arcs  subtend  equal  central 
angles  (angles  at  the  centre),  and  conversely. 

Corollary  6.  Of  two  unequal  chords  or  arcs,  the  greater 
subtends  the  greater  central  angle. 

What  figure  is  determined  by  two  parallel  chords  and  the 
chords  of  the  intercepted  arcs  ?  By  two  secants  that  inter- 
cept equal  arcs  and  the  central  normals  thereto  ? 

118.  Theorem  LVI.  —  A  central  angle  subtended  by  a  cer- 
tain arc  (or  chord)  is  double  the  peripheral  angle  subtended 
by  the  same  (or  an  equal)  arc  (or  chord)  (Fig.  77). 

Proof.  Let  ASB  be  a  central  angle,  and  APB  be  a 
peripheral  angle  (periphery  =  circumference,  the  circle 
itself),  subtended  by  the  same  arc  or  chord  AB.     Draw  the 


100  GEOMETRY.  [Th.   LVI. 

diameter  PD.  Then  the  A  ASP  and  BSP  are  isosceles 
(why?)  ;  hence  the  angle  ASD  =  2  angle  APD,  and  angle 
BSD  =  2  angle  BPD  (why  ?)  ;   hence  angle  ASB  =  2  angle 

APB.      Q.E.D. 


Fig.  'jj. 

Corollary  i .  All  peripheral  angles  subtended  by  (or  stand- 
ing on)  the  same  or  equal  chords  or  arcs  are  equal.  Hence, 
as  P  moves  round  from  A  to  B^  the  angle  APB  remains 
unchanged  in  size. 

Def.  An  angle  with  its  vertex  on  a  certain  arc,  and  its 
arms  passing  through  the  ends  of  that  arc,  is  said  to  be 
inscribed  in  that  arc.  Hence  for  an  angle  to  be  inscribed 
in  a  certain  arc,  and  for  it  to  stand  on  the  explernental  arc, 
are  equivalent. 

Corollary  2.  All  angles  inscribed  in  the  same  or  equal 
arcs  of  the  same  or  equal  circles  are  equal. 

Corollary  3.  As  the  vertex  /*  of  a  peripheral  angle  sub- 
tended by  an  arc  (or  chord)  AB^  in  passing  round  a  circle 
goes  through  either  end  of  the  arc  (or  chord),  the  angle 
itself  leaps  in  value,  changes  to  its  supplement. 


'P^  (2^^f(>y^ 


Tn.  LVIII.]  THE   CIRCLE.  101 

1 19.  Theorem  LVII.  —  The  locus  of  the  vertex  of  a  given 
angle  standing  on  a  given  tract  ii  two  symmetric  circular 
arcs  through  the  ends  of  the  tract  (Fig.  78). 

\0 


Fig.  78. 

Proof.  Let  P  be  the  vertex  of  the  given  angle,  in  any 
position,  standing  on  the  tract  AB.  Through  A,  Py  and  B 
draw  a  circular  arc  subtended  by  AB.  We  have  just  seen 
that  as  long  as  P  stays  on  this  arc,  the  angle  P  remains  the 
same  in  size.  Moreover,  the  point  P  cannot  be  without  the 
arc,  as  at  O,  because  the  angle  A  OB  is  less  than  APB 
(why?)  ;  neither  can  it  come  within  the  arc,  as  to  /,  because 
the  angle  A  IB  is  greater  than  APB  (why  ?)  ;  hence  so  long 
as  the  angle  is  constant  in  size  the  vertex  must  remain  on 
the  arc  APB  or  on  its  symmetric  arc  AP'By  of  which  plainly 
the  same  may  be  said.     q.  e.  d. 

120.  Theorem  LVIII.  —  The  angle  inscribed  in  a  semi- 
circle (or  standing  on  a  semicircle  or  diameter)  is  a  right- 
angle  (Fig.  79). 


102 


GEOMETRY. 


[Th.  LVIII. 


Proof.  Let  ABC\>^  any  angle  in  a  semicircle.  Then  it 
is  half  of  the  central  angle  ASC  (why?),  which  is  a  straight 
angle  (why?).     Q. e. d. 

U 


Fig.  79. 

Now  let  the  vertex  B,  the  intersection  of  the  rays  L  and 
N,  move  round  the  circle  toward  C;  the  angle  ABC  re- 
mains a  right  angle,  710  matter  how  close  B  approaches  to  C ; 
moreover,  when  B  passes  C,  into  the  lower  semicircle,  the 
angle  remains  a  right  angle  (why?).  That  is,  the  angle  at^ 
remains  a  right  angle,  no  matter  from  which  side  nor  how 
close  B  approaches  to  C.  Hence  it  is  a  right  angle  even 
when  B  falls  on  C.  But  then  the  ray  L  falls  on  the  diame- 
ter A  C,  hence  the  ray  N  takes  the  position  T  normal  to  the 
diameter  (or  radius)  at  its  end.  Such  a  normal  to  a  radius 
at  its  end  is  called  a  tangent  to  the  circle  at  the  point  of 
tangence  (or  touch  or  contact^  C. 

Def.  A  ray  normal  to  a  tangent  to  a  curve  at  the  point 
of  touch  is  called  normal  to  the  curve  itself.     Hence 

Corollary.  All  radii  of  a  circle  are  normal  to  the  circle ; 
and  conversely,  all  normals  to  a  circle  are  radii  of  the  circle. 


Th.  LX.] 


THE   CIRCLE. 


103 


121.  Theorem  LIX.  —  All  points  on  a  tangent y  except  the 
point  of  contact,  lie  outside  of  the  circle. 

Proof.  For  the  point  of  touch  is  distant  radius  from  the 
centre  (why?),  and  all  other  points,  as  D,  of  the  tangent 
are  further  from  the  centre  (why  ?)  ;  hence  all  other  points 
of  the  tangent  are  without  the  circle  (why?),     q.  e.  d. 

122.  Theorem  LX.  —  The  point  of  tangence  is  a  double 
point. 

Proof.  For  it  is  on  a  diameter,  or  axis  of  symmetry,  of 
the  circle,  and  every  such  point  is  a  double  point  with 
respect  to  that  axis. 

Independently  of  this  consideration,  it  is  seen  that  the 
chord  CB  becomes  the  tangent  CT  when,  and  only  when, 
the  points  B  and  C  fall  together  in  C. 


Fig.  8o. 


Still  otherwise,  let  AB  be  any  chord  of  a  circle  about 
(Fig.  8o)  O.  Draw  the  mid-normal  OD.  Now  let  the 
circle  shrink  about  the  centre  O  :  the  points  A  and  B  move 


104 


GEOMETRY. 


[Th.  LXI. 


towards  each  other,  and  as  D  is  ahvays  mid-way  between 
them  they  finally  fall  together  in  D,  and  their  join  is  tan- 
gent at  D  to  the  circle  of  radius  OD. 

Def.  Two  points  thus  falling  together  in  a  double  point 
are  called  consecutive  points.  Accordingly  we  may  define 
a  tangent  to  a  circle  (or  to  any  curve)  as  a  ray  through  two 
consecutive  points  of  the  circle  (or  curve).  Adopting  this 
definition,  let  the  student  prove 

123.  Theorem  LXI.  Every  tangent  to  a  circle  is  normal 
to  a  radius  at  its  e?id ;  conversely,  Every  normal  to  a  radius 
at  its  end  is  tangent  to  the  circle. 

124.  Theorem  LXII.  The  angle  between  a  tangent  and 
a  chord  equals  the  peripheral  angle  on  the  same  chord,  or 
equals  half  the  angle  of  the  chord  (Fig.  81). 


Proof.  For  if  Z>r  be  a  diameter,  then  the  angles  BDT 
and  BTA  are  equal,  being  complements  of  the  same  angle 
BTD  (why?).     Or  thus  :   TB '\?>  sl  chord,  and  TA  is  also  a 


Th.  LXIV.l 


THE   CIRCLE, 


105 


chord,  through  the  double  point  T\  hence  the  angle  BTA 
is  a  peripheral  angle  standing  on  the  arc  TB.     q.  e.  d. 

125.  Theorem  LXIII.  —  The  angle  behveen  two  secants 
is  half  the  sum  or  half  the  difference  of  the  angles  of  the 
intercepted  arcSy  according  as  the  secants  intersect  within  or 
without  the  circle. 

Proof.  For  on  drawing  AB^  the  angle  /  is  seen  (Fig. 
82)  to  be  the  sum,  and  the  angle  O  the  difference,  of  the 


Fig.  82. 
angles  at  A  and  B^  standing  on  the  arcs  AA'  and  BB\ 

Q.  E.  D. 

126.  Theorem  LXIV. — An  encyclic  quadrangle  has  its 
opposite  angles  supplemental. 

Proof.  For  the  angles  B  and  D  are  halves  of  the  two 
central  angles  ASC  and  CSAj  whose  sum  is  a  round  angle. 
Hence  the  sum  of  B  and  Z>  is  a  straight  angle,     g.  e.  d. 


106  GEOMETRY.  [Th.  LXV. 

127.    Theorem  LXV. — Conversely,  A    quadrangle  with 
its  opposite  angles  supplemental  is  encyclic  (Fig.  Zt^), 


Fig.  83. 

Proof.  Let  ABCD  be  the  quadrangle  with  the  angles 
A  and  C,  B  and  D,  supplemental.  About  the  A  ABC  draw 
a  circle.  If  P  be  any  point  on  the  arc  of  this  circle  exple- 
mental  to  ABC,  then  the  angle  AFC  is  the  supplement  of 
ABC ;  but  if  B  be  not  on*  this  arc,  then  the  angle  ABC  is 
either  greater  or  less  than  that  supplement  (why?).  Now 
the  angle  Z>  is  that  supplement;  hence  B>  is  on  the  arc. 

Q.  E.  D. 

128.    Relations  of  circles  to  each  other. 

"  Suppose  two  circles  X  and  K'  of  radii  r  and  r'  to  be 
concentric,  i.e.  to  have  the  same  centre  O.  Then,  plainly, 
the  distance  between  them  measured  on  any  half-axis  OR 
is  r—r\  the  difference  of  the  radii.  Draw  tangents  AT, 
A^T\  where  00^  cuts  the  circles.  They  are  parallel  (why  ?). 
Now  let  the  centre  of  K^  move  out  on  (9(9'  a  distance  r  —  r\- 
then  A  falls  on  A^  and  A'T'  on  AT;  the  circles  have  a 
common  tangent  at  A  and  are  said  to  touch  each  other 
inner ly  at  A  (Fig.  84). 


Tn.   LXVII.] 


THE   CIRCLE 


107 


Now  let  (9'  move  still  further  along  00* ;  then  the  circles 
will  lie  partly  within,  partly  without,  each  other ;  they  will 
intersect  at  two  i)oints,  and  only  two  (why?),  symmetric  as 
to  00*  (why?),  namely  /'and  P  \  hence 


Fig.  84. 

Theorem  LXVI.  —  The  common  axis  of  two  circles  is  the 
mid- normal  of  their  common  chord. 

When  6>'  is  distant  ;■  -f  / '  from  Oy  the  circles  lie  without 
each  other,  but  still  have  a  common  tangent  (why?)  and  are 
said  to  touch  outerly. 

As  O*  moves  still  further  away  from  O,  the  circles  cease  to 
touch  and  henceforth  lie  entirely  without  each  other. 

Thus  we  find  that  there  are  three  critical  positions  depend- 
ing on  the  distance  d  between  the  centres  O  and  (9' : 

d=Oy  when  the  circles  are  concentric. 

d—r—r'j  when  the  circles  touch  innerly. 

d=  r-\-  r',  when  the  circles  touch  outerly. 
There  are  alio  three  intermediate  positions  : 
For  o  <  d  <  r  —  /  the  one  circle  is  i,<iihin  the  other. 
F'or  r—r'<  d<  r  -\-  r  the  circles  intersect. 
For  r  4-  r  <  //  <  00  the  circles  lie  ivithout  each  other. 

129.  Theorem  LXVII .  —  From  a ny  point  without  a  circle 
tivOf  and  only  two^  tangents  may  be  drawn  to  the  circle  (Fig. 

85)- 


108 


GEOMETRY. 


[Th.   LXVII. 


Proof.  Let  O  be  the  centre  of  the  circle  K^  and  F  be 
the  point  without.  On  OP  as  a  diameter  draw  a  circle  K^ ; 
only  one  such  circle  is  possible  (why?),  and  it  cuts  K  in  two, 
and  only  two,  points,  T  and  T.  Draw  PT  and  PV  :  they 
are  tangent  to  ^  at  T  and  V  (why  ?) .  Moreover,  no  other 
ray  through  P,  as  PU,  is  tangent  to  K^  because  OUP  is  not 
a  right  angle  (why?).     Q.  e.  d. 


Fig.  85. 

Def.  The  chord  TT^  through  the  points  of  contact  of  the 
tangents  is  called  the  chord  of  contact  for  the  point  P  or 
the  polar  of  the  pole  P  (see  Art.      ) . 

The  angle  between  the  tangents  to  two  curves  at  the 
intersection  of  the  curves  is  called  the  angle  between  the 
curves  themselves.  When  this  is  a  right  angle,  the  curves 
are  said  to  intersect  orthogonally. 

The  distance  PT  or  PT^  is  called  the  tangent-length 
from  P  to  the  circle. 

Corollary  i .  Two  circles,  one  having  as  radius  the  tangent- 
length  from  its  centre  to  the  other,  intersect  orthogonally. 

Corollary  2.  Two  tangents  are  symmetric  as  to  the  axis 
through  their  intersection ;  hence,  also,  the  tangent-lengths 
are  equal. 


Th.  LXTX.] 


THE   CIRCLE. 


109 


130.  Theorem  LXVIII. — All  tangent-lengths  to  a  circle 
from  points  on  a  concentric  circle  are  equal,  and  intercept 
equal  arcs  0/  the  circle  (Fig.  86). 


Fig.  86. 

Proof.  For  if/*  be  any  point  without  the  circle  K\  we 
may  turn  F  round  about  the  centre  (9  on  a  concentric  circle 
AT'  without  affecting  any  of  the  relations  obtaining  (why  ?) . 

Or  thus :  the  right  A  TOP  and  T'OP^  are  plainly  con- 
gruent (why?);  hence  PT=  P^T  (why?),     q.  e.  d. 


*i3i.  Theorem  LXIX.  —  The  intercept  between  tioo  fixed 
tangents  on  a  third  tangent  subtends  a  constant  central  angle 
(Fig.  87). 

Proof.  Let  PT  and  PT  be  the  fixed  tangents,  VV^  the 
intercept  on  the  variable  ray  tangent  at  U.  Then  TPT'  is 
a  constant  angle,  and  VOV^  is  half  of  TOT  (why?),  and 
hence  is  constant.     Q.  e.  d. 


110 


GEOMETRY. 


[Tii.  LXX. 


Frc.  87. 

132.  Theorem  LXX.  — If  the  central  (or  peripheral)  angles 
of  the  common  chord  of  two  intersecting  circles  be  equal,  the 
circles  are  equal. 

Let  the  student  conduct  the  proof  suggested  by  the  figure 
(Fig.  88),  and  let  him  prove  the  converse. 


*I33.  Theorem  LXXI.  —  The  circumcircle  of  a  A  equals 
the  circumcircle  of  the  orthocentre  and  any  two  vertices  of 
the  l^  (Fig.  89). 

Proof.  Let  K  be  the  circumcircle  of  the  A  ABC,  K'  the 
circumcircle  of  A,  B,  and  O  the  orthocentre.     The  angles 


Th.  LXXII.]  the   circle.  Ill 

Cand  B'OA^  are  supplemental  (why?)  ;  also  the  angles  D 
and  BOA  are  supplemental  (why?)  ;  and  the  angles  BOA 
and  B'OA'  are  equal  (wliy?)  ;  hence  the  angles  Z>  and  C 
are  equal ;  hence  A'  =  A''  (why  ?)     q.  e.  d. 


Fig.  89. 

♦134.  Theorem  LXXII. — T/if  mid-points  of  the  sides  of  a 
A,  the  feet  of  its  a/titi/deSy  and  the  mid-points  between  its 
orthocentre  and  vertices^  are  nine  encyclic  points. 

Proof.  I^t  a  circle  through  X^  V,  Z,  the  mid-points  of  the 
sides,  cut  the  sides  in  three  other  points,  0]  F,  IV.  Then 
the  angle  ZXy=  angle  A  (why?),  and  also  =  angle  ZFV 
(why?)  ;  therefore  the  A  AZF is  symmetrical.  Hence  the 
A  Zf^B  is  also  symmetrical,  Z  is  equidistant  from  A,  F,  and 
Bf  and  the  angle  A FB  is  a.  right  angle  (why?)  ;  so  also  the 
angles  at  6^  and  IV;  i.e.  the  circle  through  the  mid-points  of 
the  sides  goes  through  the  feet  of  the  altitudes  (Fig.  90). 

Again,  if  the  circle  cuts  the  altitudes  at  /*,  Q^  R,  then  the 
angle  F/'[r=  angle  FZ/r(why?)  =  2  angle  FA IV (why?). 
Moreover,  A,  F,  O,  IV,  are  encyclic  (why?)  ;  hence  AO  is 
a  diameter  of  the  circle  through  them  (why?)  ;  and  FA  IF 
is  a  peripheral  angle  standing  on  the  arc  FIV-,  hence  the 


112 


GEOMETRY. 


[Th.   LXXIII. 


double  angle  yPJV  must  be  the  central  angle  of  the  same 
arc;  i.e.  /'is  the  mid-point  between  a  vertex  and  orthocen- 
tre  :  so,  also,  are  Q  and  /?,  similarly,     q.  e.  d. 


De/.  This  remarkable  circle  is  called  the  9-point  circle, 
or  a'r^/e  of  Feuerbach,  of  the  A  ABC. 

Corollary.  The  radius  of  the  9-point  circle  is  half  the 
radius  of  the  circumcircle. 

135-  Def.  A  Polygon  all  of  whose  sides  touch  a  circle  is 
said  to  be  circumscribed  about  it,  and  the  circle  is  said  to  be 
inscribed  in  the  polygon. 

Theorem  LXXIII.  — A  circle  may  be  inscribed  in  any  A. 

Proof.  Let  ABC  be  any  A  (see  Fig.  59).  Draw  the 
inner  mid-rays  of  the  angles  at  A,  B,  C ;  they  concur  in  the 
in-centre  /of  the  A,  equidistant  from  the  three  sides  (why?). 
About  this  point  as  centre  with  this  common  distance  as 
radius  draw  a  circle ;  it  will  touch  the  three  sides  of  the  A 
(why  and  where  ?) .     Q.  e.  d. 

N.B.  We  have  seen  that  the  outer  mid-rays  of  the  angles 
concur  in  pairs  with  the  inner  mid-rays  of  the  angles  in  the 
three  ex-centres  Ei,  Eo,  E^,  also  equidistant  from  the  sides 


Th.  LXXIV.] 


THE   CIRCLE. 


113 


(Fig.  60).  The  circles  about  these  touch  only  two  sides 
innerly,  but  the  third  side  outerly,  and  hence  are  called 
escribed,  or  ex-circles. 

Corollary.  Four,  and  only  four,  circles  touch,  each,  all  the 
sides  of  a  A. 

135  a.  Theorem  LXXIV.  — ///  a  4- side  circumscribed 
about  a  circle  the  sums  of  the  tivo  pairs  of  opposite  sides  are 
equal  (Fig.  91). 


Fig.  91. 

Proof.     The  sum  of  the  four  sides  is  plainly  2t-{-  2u-\-  2v 
-f  20/,  and  the  sum  of  either  pair  of  opposites  is  t-\-u-\-v^'w. 

Q.  E.  D. 

Conversely,  If  the  sums  of  two  pairs  of  opposite  sides  of  a 
4-side  be  equals  the  4-side  is  circumscribed  about  a  circle. 

Proof.     Let  two  counter  sides,  AB  and  DC  meet  in  /, 
and  inscribe  a  circle  K  in  the  triangle  ADI.    Through  B 


114 


GEOMETRY. 


[Th.  LXXV. 


draw  a  tangent  (Fig.  92)  to  A' at  U,  and  let  it  cut  DI 2X  C. 
Then  since  ABCD  is  circumscribed  about  K,  we  have 


or 


Fig.  92. 

AB^CD^BC  ^DA, 
Also  AB+CD  ==BC  +  nA  (why ?) . 

Whence  CD-CD  =  BC-B  C, 

CC=^BC-BC. 


Hence  C  and  C  fall  together  (why?     Art.  56).     q.  e.  d. 

136.  Theorem  LXXV.  —  The  tan  gent- length  from  a  ver- 
tex of  a  A  to  the  in-circle  equals  half  the  perimeter  of  the  A 
less  the  opposite  side  (Fig.  93). 


Th.   LXXVI.] 


THE   CIRCLE. 


lis 


Proof.  For  the  sum  of  CE  ^  CD -\-  BD  +  BF  is  plainly 
2a  (why?)  ;  subtract  this  from  the  whole  perimeter,  a  -\-  b  -\-  c, 
and  there  remains  AE  -\-  AF=  a  -\- d  +  c  —  la^  or  AE — 


d  +  c- 


=^AF. 


FIG.  93. 

It  is  common  and  convenient  to  denote  the  perimeter 
( Fig.  93)  ( =  measure  round  =  sum  of  sides)  by  2s \  then 
we  see  that  the  tangent-lengths  from  Ay  B,  C,  are  s  —  a^ 
s  —  b,  s  —  c. 

Corollary.  The  tangent-length  from  any  vertex,  A,  of  a 
A  to  the  opposite  ex-circle  and  the  two  adjacent  ex-circles 
are  s,  s—by  s  —  c.  Hence  s  —  a,  s^  s  —  b,  s  —  c,  are  the 
four  tangent-lengths  from  any  vertex,  ^,  of  a  A  to  the  in- 
circle  and  the  three  ex-circles. 

These  relations  are  useful  and  important. 

137.    Theorem  LXXVI.  —  There  is  a  regular  n-side. 
Proof.     For  the  angle  is  a  continuous  magnitude  (why?)  ; 
hence  there  are  angles  of  all  sizes  from  zero  to  a  round 


116 


GEOMETRY. 


[Th.  LXXVI. 


angle  ;    hence  there  is  an  angle,  the  -  part  of  a  round  angle, 

such  that,  taken  71  times  in  addition,  the  sum  will  be  a  round 
angle.  Suppose  such  an  angle  drawn,  whether  or  not  we 
can  actually  draw  it,  and  suppose  ;/  such  angles  placed  con- 
secutively around  any  point  O,  so  as  to  make  a  round  angle. 
In  other  words,  suppose  n  half-rays  drawn  cutting  the 
round  angle  about  O  into  n  equal  angles.  Draw  a  circle 
about  (9,  with  (Fig.  94)  any  radius,  and  draw  the  ;/  chords 


Fig.  94. 

subtending  the  n  equal  central  angles.  These  chords  are 
all  equal  (why?),  and  subtend  equal  arcs,  and  they  form  an 
?z-side.     Moreover,  the  angle  between  two  consecutive  sides 


Th.  LXXVIIL]  the   circle.  117 

is  constant  in  size,  because  it  stands  on  the  part  of 

n 

the  circle.     Hence  the  //-side  is  both  equilateral  and  equian- 
gular ;  that  is,  it  is  regular,     q.  e.  d. 

Corollary.     The   inner  angle  of  a  regular  w-side  is  the 

part  of  a  straight  angle. 


(^■) 


Find  the  value  in  degrees  of  the  inner  angles  of  the  first 
ten  regular  «-sides. 

N.B.  The  foregoing  demonstration  merely  settles  the 
question  of  the  existence  or  lo^^ical possibility  of  the  regular 
w-side.  The  problem  of  actually  drawing  such  a  figure  is 
one  of  the  most  intricate  in  all  mathematics,  and  has  been 
solved  only  for  certain  very  special  classes  of  values  of  n. 
Hut  in  order  to  discover  the  properties  of  the  figure,  it  is  by 
no  means  necessary  to  be  able  to  draw  it  accurately.  It  is 
only  since  1864  that  we  have  known  how  to  draw  a  straight 
line  or  ray  exactly. 

137a.  Theorem  LXXVII.  —  The  vertices  of  a  regular 
n-side  are  encyclic  (Fig.  94). 

Proof.  Through  any  three  vertices,  as  A,  B,  C,  of  a  regular 
«-side,  draw  a  circle  K ;  about  C  with  radius  CB  draw 
another  circle.  The  fourth  vertex  D  must  lie  on  this  circle 
(why?).  If  it  lie  on  the  circle  A',  then  the  angle  BCZ>  = 
angle  ABC,  as  is  the  case  in  the  regular  «-side.  Neither 
can  it  lie  off  of  X,  as  at  D'  or  Z>",  because  then  the  angle 
BCD'  or  BCD"  would  not  equal  angle  BCD  (why?),  and 
hence  would  not  equal  angle  ABC.  Hence  the  next  vertex 
must  lie  on  the  same  circle  K,  and  so  on  all  around,    q.  e.  d. 

138.  Theorem  LXXVIII.  —  T/te  sides  of  a  regular  n-side 
are pericyclic  (that  is,  they  all  touch  a  circle). 


118 


GEOMETRY. 


[Th.  LXXIX. 


Proof.    For,  on  drawing  the  radii  of  the  circumcircle  K 
(Fig.  95)  to  the  vertices,  we  get  n  congruent  symmetric  A 
A 


O  Fig.  95. 

(why?).  The  altitudes  of  all  are  the  same  (why?)  ;  with 
this  common  altitude  as  radius  draw  another  circle,  K\ 
about  the  same  centre.  It  will  touch  each  of  the  sides 
(why?).     Q.  E.  D. 

Corollary.  The  points  of  touch  of  the  sides  of  the  regular 
circumscribed  ;/-side  are  mid-points  of  the  sides. 

139.  Theorem  LXXIX.  —  The  points  of  touch  of  a  regular 
circujnscj'ibed  11- side  are  the  vertices  of  a  regular  inscribed 
n-side. 

Proof.  Connect  the  points  of  touch  consecutively.  Then 
the  A  so  formed  are  all  congruent  (why?)  ;  hence  the 
joining  chords  are  equal ;  hence  the  arcs  are  equal ; 
hence  the  Theorem,     q.  e.  d. 


CIRCLE  AS  ENVELOPE,  119 

THE   CIRCLE  AS   ENVELOPE. 

*i4oa.  Thus  far  we  have  regarded  the  circle  from  various 
points  of  view ;  from  the  most  familiar  it  was  seen  to  be  the 
locus  of  a  point  in  a  plane  at  a  fixed  distance  from  a  fixed 
point.  An  almost  equally  important  conception  of  the  curve 
treats  it  not  as  the  locus  of  a  point,  but  as  the  envelope  of  a 
ray.  If  the  point  P  moves  in  the  plane  always  equidistant 
from  O,  then  its  locus  is  the  circle,  on  which  it  may  always 
be  found  ;  also,  if  the  ray  R  moves  about  in  the  plane  always 
equidistant  from  O^  then  its  etwelope  is  the  circle,  on  which 
it  may  always  be  found,  on  which  it  lies,  which  it  continually 
touches.  The  point  traces  the  circle,  the  ray  envelops  the 
circle,  which  is  accordingly  called  the  envelope  (/>.  the 
enveloped  curve  —  French  enveloppee)  of  the  ray.  In  higher 
mathematics  the  notion  of  the  ray,  instead  of  the  point,  as 
the  determining  element  in  the  nature  of  a  curve,  attains 
more  and  more  significance.  In  this  text  we  are  confined 
to  the  circle  —  the  envelope  of  a  ray  in  a  plane,  at  a  fixed 
distance  from  a  fixed  point. 

*i40b.  It  is  not  only  rays,  however,  that  may  envelop  a 
curve ;  but  circles,  and  in  fact  any  other  curves.  Thus,  let 
the  student  draw  a  system  of  equal  circles,  having  their 
centres  on  another  circle ;  the  envelope  will  at  once  be  seen 
to  be  a  pair  of  concentric  circles.  Let  him  also  find  the 
envelope  of  a  system  of  circles  equal  and  with  centres  on  a 
given  ray.  In  general,  let  him  find  the  envelope  of  a  circle 
whose  centre  moves  on  any  given  curve.  Lastly,  let  him 
draw  a  large  number  of  circles  all  of  which  pass  through  a 
fixed  point,  while  their  centres  all  lie  on  a  fixed  circle,  and 
let  him  observe  what  curve  they  shadow  forth  as  envelope. 

Show  that  as  the  pole  of  a  chord  (or  ray)  traces  a  circle, 


120  GEOMETRY. 

the  chord  itself  envelops  a  concentric  circle,  and  con- 
versely. 

Show  that  tangents  from  two  points  on  a  centre  ray  form 
a  kite,  and  conversely.  Also  the  chords  of  contact  are 
parallel,  and  conversely. 

O  is  the  centre  of  a  circle,  P  any  point  without  it.  Show 
how  to  find  the  point  of  touch  of  the  tangents  from  /*,  by 
drawing  a  circle  about  O  through  P  and  a  tangent  where 
OP  cuts  the  given  circle. 

CONSTRUCTIONS. 

140.  Hitherto,  in  our  reasoning  about  concepts,  figures 
have  not  been  at  all  necessary,  though  exceedingly  useful  in 
making  sharp  and  precise  our  imagination  of  the  relations 
under  consideration,  in  furnishing  sensible  examples  of  the 
highly  general  notions  that  we  dealt  with.  The  conclusions 
reached  thus  far  all  he  wrapt  up  in  axioms  and  in  our  defi- 
nitions of  point,  ray,  and  circle,  and  our  work  has  been  one 
of  explication  only ;  we  have  merely  brought  them  forth  to 
light.  Our  demonstrations  have  not  presumed  ability  to  draw 
accurately,  and  would  remain  unshaken  if  we  could  not  draw 
at  all.  Nevertheless,  for  many  practical  purposes,  it  is  ex- 
tremely important  and  even  indispensable  that  we  actually 
make  the  constructions  and  draw  the  figures  that  thus  far  we 
have  merely  supposed  made  and  drawn. 

141.  What  is  meant  by  drawing  a  ray,  circle,  or  any  line? 
Any  mark,  whether  of  ink  or  chalk,  though  a  solid,  may  be 
treated  as  a  line  by  abstraction.  Only  its  length,  not  its 
width  nor  thickness,  concerns  us.  How  to  make  not  just 
any  mark,  but  some  particular  mark  called  for,  is  our  prob- 
lem  (7rpoft\7)ixa  =  anything  thrown  forward  as  a  task),  and 


CONS  TR  UC  TIONS.  1 2 1 

its  solution  consists  accordingly  of  two  parts,  the  logical  and 
the  mechanical.  The  first  is  accomplished  by  fixing  exactly 
in  thought  the  position  of  all  the  geometric  elements  (points, 
rays,  circles)  in  question  ;  the  second,  by  making  marks  that 
by  abstraction  may  be  treated  as  these  elements.  Now,  a 
point  is  fixed  as  the  join  of  two  rays,  a  ray  as  the  join  of  two 
points  (by  what  axiom?)  ;  a  circle  is  fixed  or  determined  by 
its  centre  and  radius  (why?),  or  by  three  points  on  it  (why?). 
Accordingly,  when  we  know  two  rays  through  a  point,  or  two 
points  on  a  ray,  or  centre  and  radius,  or  three  points  of  a 
circle,  we  know  the  point,  or  ray,  or  circle  completely.  The 
logical  part  of  our  work  is  finished,  then,  when  we  determine 
every  point  as  the  join  of  two  known  rays,  every  ray  as  the 
join  of  two  known  points,  every  circle  as  drawn  through  three 
known  points  or  about  a  known  centre  with  a  known  radius. 
The  mechanical  part  of  the  solution  requires  us  to  put  and 
keep  a  point  in  motion  along  a  circle  or  a  ray.  Circular 
motion  is  brought  about  by  the  compasses  already  described 
(Art.  114),  of  which  the  shape  is  arbitrary,  the  necessary 
parts  being  merely  a  fixed  point  rigidly  connected  in  any  way 
with  a  movable  point.  But  in  the  ruler  one  edge  is  supposed 
made  straight  to  begin  with,  so  that  a  pencil  point  gliding 
along  it  may  trace  a  straight  mark.  Hence  the  nse  of  the 
ruler  is  really  illogical,  since  it  assumes  the  problem  of  draw- 
ing a  ray  or  straight  line  as  already  solved  in  constructing 
the  straight  edge.  To  say  that,  in  order  to  draw  a  straight 
liney  we  must  take  a  straight  edge  and  pass  a  pencil  point 
along  it,  is  no  better  logically  than  to  say  that,  in  order  to 
draw  a  circular  line,  we  must  take  a  circular  edge  and  pass 
a  pencil  point  along  it.  The  question  at  once  arises.  How 
make  the  edge  straight  or  circular  in  the  first  place  ?  It  was 
not  until  1864  that  Peaucellier  won,  though  he  did  not  at 
once  receive,  the  Montyon  prize  from  the  French  Academy 


122  GEOMETRY. 

by  solving  the  thousand-year-old  problem  of  imparting  rec- 
tilinear motion  to  a  point  without  guiding  edge  of  any  kind 
(Page  ooo).  But,  though  the  ruler  is  logically  valueless,  it  is 
practically  invaluable,  even  after  the  great  discovery  of  Peau- 
cellier.  Its  edge  being  assumed  as  straight  and  of  any 
desired  length,  and  a  pair  of  compasses  of  adjustable  size 
being  given,  we  now  make  the  following  Postulates  : 

I.  About  any  point  may  be  drawn  a  circle  of  any  radius. 

II.  Through  any  tivo  points  may  be  drawn  a  ray  (more 
strictly,  a  tract  of  any  required  length) . 

Corollary.  On  any  ray  from  any  point  on  it  we  may  lay 
off  a  tract  of  any  required  length. 

These  are  the  only  instruments  used  or  postulates  assumed 
in  the  constructions  of  Elementary  Geometry. 

142.  The  fundamental  relations  of  rays  to  each  other  are 
two  :  Normality  and  Parallelism.     Hence 

Problem  I.  —  To  draw  a  ray  normal  to  a  given  ray.  Since 
there  are  many  rays  normal  to  a  given  ray,  to  make  the 
problem  definite  we  insert  the  limiting  condition,  through  a 
given  point.     Two  cases  then  arise  : 

A.  When  the  given  point  is  on  the  given  ray.  All  we  can 
do  is  to  draw  a  circle  about  the  point  P.  It  cuts  the  ray  at 
two  points,  A  and  A\  symmetric  as  to  P.  Hence  the  mid- 
normal  of  AA^  is  the  normal  sought.  Hence  any  point  on 
this  normal  lies  on  two  circles  of  equal  radius  about  A  and 
A\     Hence  (Fig.  96) 

Solution.  From  the  given  point  P  lay  off  on  the  given 
ray  two  equal  tracts  PA,  PA\  About  A  and  A^  draw  two 
equal  circles.  Through  their  points  of  intersection  draw 
their  common  chord.     It  is  the  normal  sought. 


CONS  7'A' re  IVOiVS. 


123 


Proof.  For  it  is  the  mid -normal  of  A  A',  since  it  has  two 
of  its  i)oints  equidistant  from  //  and  A',  and  P  is  the  mid- 
point of  A  A'. 


Fig.  96. 


Query :  What  radius  shall  we  take  for  the  circles  about 
A  and^'? 

B.  IV/irn  the  given  point  is  not  on  the  given  ray.  All  we 
can  do  is  to  draw  a  circle  about  the  given  point  P.  Let  it 
cut  the  ray  at  A  and  A\  Then  the  mid-normal  to  AA^  is 
the  normal  required  (why?).     Hence  (Fig.  97) 

Solution.  Determine  the  points  A,  A'  on  the  ray  by 
a  circle  about  the  given  point  F;  then  proceed  as  in  the 
first  case  (A). 


124 


GEOMETRY. 


Proof.     For    the    mid-normal    of  AA   goes   through   P 
(why?). 


P 


\ 


A' 


"Ta 


Fig.  97. 
Query :    What  radius  shall  we  take  for  the  circle  about  P  ? 

143.  Problem  II.  —  To  draw  a  parallel  to  a  given  ray. 
Since  there  are  many  parallels  to  every  ray,  to  make  the 
problem  definite  we  must  insert  the  limiting  condition, 
through  a  given  point ;  then  it  becomes  perfectly  definite 
(why?).  Manifestly  the  point  must  be  not  on  the  ray 
(why?).  We  now  reflect  that  a  transversal  makes  equal 
corresponding  angles  with  parallels,  and  we  have  just  learned 
to  draw  a  normal  transversal.     Hence  (Fig.  98) 

Solution.  Through  the  point  draw  a  normal  to  the  ray ; 
through  the  same  point  draw  a  normal  to  this  normal.  It 
will  be  the  parallel  required. 

Proof.  For  it  goes  through  the  point  and  is  parallel  to  the 
ray  (why?). 

These  two  problems  have  been  discussed  at  such  length 
as  being  the  hinges  on  which  nearly  all  others  turn.      At 


CONSTRUCTIONS.  125 

the  end  of  a  problem  is  sometimes  written  q.  e.  f.  =  quod 
erat  facicTnium  —  ivhich  ivas  to  he  done,  and  translates  the 
Euclidean  h-ntp  iBei  irpa^ai. 


A'\  ;  ^A 


Fir,.  98. 

144.  Problem  III.  —  To  bisect  a  given  tracts  or  to  draw 
the  mid-normal  to  a  given  tract ,  AB, 

Proceed  as  in  Problem  I. 

145.  Problem  IV.  —  To  bisect  a  given  angle. 

Solution.  About  the  vertex  draw  any  circle  cutting  the 
arms  at  A  and  A\  and  draw  the  mid-normal  of  AA\  It  is 
the  mid-ray  sought  (why?). 

Corollary.   Show  how  to  bisect  any  circular  arc  AB, 

146.  Problem  V.  —  To  bisect  the  angle  between  two  rays 
whose  Join  is  not  given  (Fig.  99). 


126 


GEOMETRY. 


We  reflect  that  the  join  AA^  of  two  corresponding  points 
on  the  rays  makes  equal  angles  with  the  two  rays  that  form 
the  angle.     Hence 

Solution.  From  any  point  P  oi  L  draw  the  normal  to  it, 
cutting  M  at  Q.     From  Q  draw  the  normal  to  M.     Bisect 


Fig.  99. 

the  angle  at  Q  between  these  two  normals  by  the  mid-tract 
QR.  Draw  the  mid-normal  of  QR.  It  is  the  mid-ray 
sought  (why?). 

147.    Problem  VI.  —  To  multisect  a  given  tract  AB  (Fig. 
100). 

L 


CONSTRUCTIONS, 


127 


Solution.  Through  either  end  of  the  tract,  as  A,  draw  any 
ray,  and  lay  off  on  it  from  A  successively  n  equal  tracts,  L 
being  the  end  of  the  last.  Draw  BL.  Through  the  ends 
of  the  equal  tracts  draw  parallels  to  BL.  They  cut  AB 
into  n  equal  parts  (why?). 

148.  Problem  VII.  —  To  draw  an  angle  of  given  size^  i.e. 
equal  to  a  given  angle  (Fig.  loi). 


Solution.  At  any  point  A  of  either  arm  of  the  given  angle 
O  erect  a  normal  to  OA  cutting  the  other  arm  at  B.  From 
any  point  O  on  any  other  ray  lay  off  O' A*  =  OA^  and  normal 
to  the  ray  erect  A^B'  =  AB  and  draw  0^B\  Then  angle  O* 
=  angle  O  (why?). 

When  does  this  construction  fail  ?     How  proceed  then  ? 

149.  Problem  VIII.  —  To  draw  a  tiiu-t  cf  given  length 
subtending  a  given  angle  and  parallel  to  a  given    ay. 

Data:  O  the  given  angle,  L  the  ray,  a  the  length  (Fig. 
102). 

Solution.  Through  any  point  P  of  either  arm  of  the  angle 
draw  a  parallel  to  the  ray,  and  lay  off  on  it  towards  the  other 


128 


GEOMETKY. 


arm  a  tract  PA  of  the  given  length  a.  Through  A  draw  a 
parallel  to  OP,  cutting  the  other  angle  arm  at  Q ;  through  Q 
draw  a  parallel  to  PA  meeting  (9/* at  P.  QR  is  the  subtense 
sought  (why?). 


Fig.  I02. 

150.    Problem  IX.  — To  construct  a  A  : 
A.    When  altertiate  parts  (three  sides  or  three  angles)  are 
given. 

Solution.  About  the  ends  of  one  side  AB,  with  the  other 
sides  for  radii,  draw  circles  meeting  in  C.  Then  ABC  is 
the  A  sought  (why?)  (Fig.  103). 


Fig.  103. 

How  many  such  A  may  be  drawn  on  the  same  base  AB'> 
How  are  they  related?     When  is  the  solution  impossible? 


CONSTH  UCT/OiVS. 


129 


When  the  angles  are  given,  apply  the  construction  of  Problem 
VII.     How  many  solutions  arc  possible  ?     What  kind  of  A  ? 

B.  When  three  eonsecutive  parts  (two  sides  and  included 
angle  or  two  angles  and  included  side)  are  given. 

Solution.    Ai)[)ly  the  construction  in  Problem  VII. 

C.  When  opposite  parts  (two  angles  and  an  opposite  side 
or  two  sides  and  an  opposite  angle)  are  given. 

Solution.  Apply  the  constniction  in  Problem  VII.  When 
is  the  construction  ambiguous  ? 

D.  When  two  sides  and  the  altitude  to  the  third  side  are 
given. 

Solution.  Through  one  end  of  the  altitude  draw  a  normal 
to  it  for  the  base  ;  about  the  other  end  C  as  centre,  with  the 
sides  as  radii,  draw  circles  cutting  the  base  at  A  and  A\ 
B  and  B' ;  then  ACB  or  A'CB'  is  the  A  required.     Why? 

E.  When  two  sides  and  the.  medial  of  the  third  side  are 
given. 

A 


Fig.  104. 


130  GEOMETRY. 

\i SA  be  the  medial  of  BC,  and  SA^  be  symmetric  with 
(Fig.  104)  SA  as  to  S,  then  ABA'C  is  a  parallelogram 
(why?)  ;  hence 

Solution.  Take  a  tract  the  double  of  the  medial.  About 
its  ends  as  centres  with  the  sides  as  radii  draw  circles  and 
then  complete  the  construction.  How  many  A  fulfilling  the 
conditions  are  possible  ?     How  are  they  related  ? 

F.    When  the  three  medials  are  given  (Fig.  105). 


Solution.  Remember  that  the  medials  trisect  each  other ; 
construct  the  A  OBC  according  to  (E),  and  draw  OA 
counter  to  OM  and  twice  as  long. 

151.  Problfem  X.  —  To  construct  an  angle  of  given  size 
and  subtended  by  a  given  tract. 

Data :    O  the  given  angle,  AB  the  given  tract  (Fig.  106). 

Solution.  Construct  the  angle  BAD  of  given  size  (how?), 
draw  the  mid- normal  of  AB,  meeting  AD  at  P)  also  the 
normal  to  AD  at  A,  meeting  the  mid-normal  at  S.     About 


CONS  TR  UC  TIONS. 


131 


^S"  as  a  centre  with  radius  SA  draw  a  circle  ;  it  touches  AD 
at  A  (why?).  The  vertex  V  of  the  required  angle  may  be 
anywhere  on  the  arc  A  VB  or  on  its  symmetric  A  V^B  (why  ?) . 


p 

/ 

— "^ 

"^ 

\ 

V 

V 

Fio.  io6. 


152.  Problem  XI.  —  To  draw  a  circle  tangent  to  a  given 
ray. 

Solution.  About  any  point  S  with  a  radius  equal  to  the 
distance  of  S  from  the  ray,  Z,  draw  a  circle ;  it  will  be  a 
circle  required  (why?).  If  the  circle  must  touch  the  ray  L 
at  a  given  point  Py  then  S  must  be  taken  on  the  normal  to 
Z  through  P  (why  ?) .  If,  besides,  the  circle  must  go  through 
a  given  point  Q,  then  S  must  also  be  on  the  mid-normal  of 
y^^  (why?).     Hence  the  construction. 


132     ■  GEOMETRY. 

153.  Problem  XII.  —  To  draw  a  circle  touching  two  given 
rays. 

The  centre  may  be  anywhere  on  either  mid-ray  (why  ?) . 
If  now  the  circle  is  to  touch  a  third  given  ray,  the  centre 
must  be  also  on  another  mid-ray ;  that  is,  it  must  be  the 
intersection  of  two  mid-rays  of  the  three  angles  of  the  three 
rays.  There  are  four  such  intersections  —  what  are  they  ? 
Complete  the  construction.     See  Fig.  60. 

154.  Problem  XIII.  —  To  draw  a  circle  through  two 
points. 

The  centre  6"  may  be  anywhere  on  the  mid-normal  of  the 
tract  AB  between  the  points  (why?),  the  radius  is  —  what? 
If  now  the  circle  is  to  pass  through  a  third  point  C,  then  *S 
must  also  be  on  the  mid-normal  oi  BC  and  CA  (why?). 
There  is  one,  and  only  one,  such  point  (why  ?)  ;  complete 
the  construction.     When  is  the  construction  impossible  ? 

155.  Problem  XIV.  —  To  draw  a  circle  through  two  given 
points  and  tangent  to  a  given  ray ;  or,  tangent  to  two  given 
rays  and  through  a  given  point. 

This  double  problem  is  mentioned  here  because  it  must 
naturally  present  itself  to  the  mind  of  the  student ;  but  the 
solution  involves  deeper  relations  than  we  have  yet  explored. 
See  Art.  000. 

Several  of  the  foregoing  problems  were  indefinite,  admit- 
ting any  number  of  solutions  :  these  latter  taken  all  together 
form  a  system  or  family.  Problems  concerning  parallelo- 
grams and  other  4-sides  may  often  be  solved  on  cutting  the 
4-side  into  two  A. 

156.  Problem  XV.  —  To  inscribe  a  regular  4-side  (square) 
in  a  circle  (Fig,  107). 


CONSTRUCTIONS,  133 

Solution.  Join  consecutively  the  ends  of  two  conjugate 
diameters.  The  4-side  formed  is  inscribed  (why?)  and  is 
a  square  (why?). 


Fig.  107. 

157.    Problem  XVI.  —  To  inscribe  a  regular  6-side  in  a 

circle. 

Solution.  Apply  the  radius  six  times  consecutively  as  a 
chord  to  the  circle  (Fig.  108).  The  figure  formed  will 
be  the  regular  6-side  (why?). 


Fig.  108. 


N.B.  This  seems  to  have  been  one  of  the  first  geometric 
problems  ever  solved.  The  Babylonians  discovered  that 
six  radii  thus  applied  would  compass  the  circle,  and  having 


134 


GEOMETRY. 


already  divided  the  circle  into  360  steps,  they  accordingly 
divided  this  number  by  6  and  thus  obtained  60  as  the 
basis  of  the  famous  sexagesimal  notation,  which  long  domi- 
nated mathematics  and  still  maintains  its  authority  un- 
diminished in  astronomy  and  chronometry. 

In  more  difficult  problems  it  is  often  advisable,  or  in  fact 
necessary,  to  suppose  the  problem  solved,  the  construction 
made,  and  investigate  the  relations  thus  brought  to  light. 
Then  the  facts  thus  discovered  may  be  used  regressively 
in  making  the  construction  required.  This  method  is  illus- 
trated in  the  following : 

158.  Problem  XVII.  —  To  draw  a  square  with  each  of 
its  sides  through  a  given  point. 

Let  A^  B,  C,  D,  be  the  four  given  points,  and  suppose 
(Fig.  109)  PQRS  to  be  the  square  properly  drawn.     Draw 


p 

c 

\ 

"g 

\            / 

?K/ 

s 

7 

i      R~' 

1 
1 

Fig.  109. 

AB,  cutting  a  side  of  the  square,  and  through  B  draw  BE 
parallel  to  the  side  cut.  Through  a  third  point  C  draw  a 
normal  to  AB,  meeting  QR  in  F.  Also  draw  FG  parallel 
to  PQ.    Then  the  A  ABE  and  CFG  are  congruent  (why  ?) . 


cuA\^/A'UC7V0jVS. 


135 


Hence  we  discover  that  CF=  AB,    This  fact  is  the  key  to 

the 

Solution.  Join  two  points  A  and  B ;  from  a  third,  C,  lay 
off  CF  equal  and  normal  to  AB,  The  join  of  Z>,  the  fourth 
point,  and  /'is  one  side  of  the  square  in  position  (why?). 
Let  the  student  complete  the  construction  and  show  that 
four  squares  are  possible. 

159.    Problem  XVIII.  —  To  trisect  a  given  angle. 

Suppose  the  problem  solved  and  the  ray  OT  to  make 
^  TOB  =  2TOA  (Fig.  no). 


Fig.  I 10. 

From  any  point  A  of  the  one  end  of  the  angle  draw  a 
parallel  and  a  normal  to  the  other  end  ;  also  draw  to  the 
trisector  a  tract  AS—  OA.     Then  the  following  relations  are 

evident : 

^  AOS^  ^  ASO=:^  SAT-\-  ^  STA  ; 

but  :fAOS=z2^  TOB  =  2  STA  ; 

hence  ^  STA  ^"^  SAT,  and  ST=  SA. 


136  GEOMETRY. 

Again,  ^  SAR  =  ^  SRA,  being  complements  of  equal 
angles  ;  hence   SA  =  SR,    TR  =  2  OA.     Hence 

Solution.  From  any  point  A  of  either  arm  of  the  given 
angle  draw  a  parallel  and  a  normal  to  the  other  arm ;  then, 
with  one  point  of  a  straight-edge  fixed  at  the  vertex  O,  turn 
the  edge  until  the  intercept  between  the  normal  and  the 
parallel  equals  2  OA.  But  to  do  this  we  need  a  graduated 
edge,  or  a  sliding  length  2  OA  on  the  edge  itself.  Accord- 
ingly, this  construction,  while  simple,  useful,  and  interesting, 
is  not  elementary  geometric  in  the  sense  already  defined. 
To  discover  such  a  solution  for  this  famous  problem,  has  up 
to  this  time  baffled  the  utmost  efforts  of  mathematicians. 

EXERCISES    II. 

1.  State  and  prove  the  reciprocals  of  Exercises  9,  10, 
II,  page  71. 

2.  Find  a  point  on  a  given  ray,  the  sum  of  whose  dis- 
tances from  two  fixed  points  is  a  minimum. 

3.  The  same  as  the  foregoing,  difference  supplacing 
sum. 

4.  A  and  A\  B  and  B\  C  and  C\  are  symmetric  as  to 
MN.     Show  that  AABC=AA'B' C. 

5.  The  inner  and  outer  mid-rays  of  the  basal  angles  of 
a  symmetric  A  form  a  kite. 

6.  The  inner  mid-rays  of  the  angles  of  a  trapezium  form 
a  kite  with  two  right  angles. 

7.  The  joins,  of  the  mid-points  of  the  parallel  sides  of  an 
anti-parallelogram,  with  the  opposite  vertices,  form  a  kite. 

8.  The  mid-rays  of  the  angles  at  the  ends  of  the  trans- 
verse axis  of  a  kite  cut  the  sides  in  the  vertices  of  an  anti- 
parallelogram. 


EXERCISES  II.  137 

9.  How  must  a  billiard  ball  be  stnirk  so  as  to  rebound 
from  the  four  sides  of  a  table  and  return  through  its  original 
place  ? 

10.  Trace  a  ray  of  light  from  a  focus  P^  to  another  given 
point  Qi  reflected  from  a  convex  polygonal  mirror. 

11.  A  ray  of  light  falls  on  a  mirror  J/,  is  reflected  along 
5  to  a  second  mirror  M\  is  thence  reflected  along  T.  Re- 
membering that  the  angle  of  incidence  equals  the  angle  of 
reflection,  show  that  the  angle  between  the  original  ray  R 
and  its  last  reflection  T  is  twice  the  angle  between  the 
mirrors  (angle  RT=  2  angle  MM').  On  this  theorem  is 
grounded  the  use  of  the  sextant. 

12.  Two  mirrors  stand  on  a  plane  and  form  an  inner 
angle  of  60° ;  a  luminous  point  P\s  on  the  mid-ray  of  this 
angle  (or  anywhere  within  it)  ;  how  many  images  of  /*are 
formed?  How  are  they  placed?  What  if  the  angle  of  the 
mirrors  be  i/«  of  a  round  angle? 

This  theorem  is  beautifully  illustrated  in  the  kaleidoscope. 

13.  A  regular  «-side  has  n  axes  of  symmetry  concurring 
in  the  centre  of  the  «-side,  which  centre  is  equidistant  from 
the  sides  of  the  //-side. 

14.  How  do  these  axes  lie  when  n  is  even?  when  n  is 
odd?  Show  that  if  //  be  even,  the  centre  is  a  centre  of 
symmetry. 

1 5 .  The  half- rays  from  centre  to  vertices  of  a  regular  «-side 
form  a  regular  pencil  of  n  half-rays,  and  those  from  the  cen- 
tre normal  to  the  sides,  another  regular  pencil ;  also  the  half- 
rays  of  each  p>encil  bisect  the  angles  of  the  other. 

16.  In  a  figure  with  two  rectangular  axes  of  symmetry 
each  point,  with  three  others,  determines  a  rectangle,  and 
each  ray,  with  three  others,  a  rhombus. 


138  GEOMETRY. 

1 7.  Find  the  axes  of  symmetry  of  two  given  tracts. 

18.  A  regular  A,  along  with  the  figure  symmetric  with  it 
as  to  its  centre,  determines  a  regular  6-angle  (6-pointed 
star) . 

19.  Two  congruent  squares,  the  diagonals  of  one  lying  on 
the  mid-parallels  of  the  other,  form  a  regular  8-angle ;  also 
find  the  lengths  of  the  intercepts  at  the  corners. 

20.  The  outer  angle  of  a  regular  «-side  is  in  times  the 
outer  angle  of  a  regular  ;//«-side. 

EXERCISES    III. 

1.  A  circle  with  its  centre  on  the  mid-ray  of  an  angle 
makes  equal  intercepts  on  its  arms. 

2.  Tangents  parallel  to  a  chord  bisect  the  subtended 
arcs,  and  conversely. 

3.  Tangents  at  the  end  of  a  diameter  are  parallel. 

4.  A  and  B  are  ends  of  a  diameter,  C  and  D  any  other 
two  points  of  a  circle ;  iS"  is  on  the  diameter,  and  angle 
AED  —  2  angle  CAD ;  find  the  possible  positions  of  £. 

5 .  From  n  points  are  drawn  2  n  equal  tangent-lengths  to 
a  circle  ;  where  do  the  points  lie  ? 

6.  In  a  circumscribed  hexagon,  or  any  circumscribed 
2  n-side,  the  sums  of  the  alternate  sides  are  equal. 

7.  If  the  vertices  of  a  circumscribed  quadrangle,  hexagon, 
or  any  2  «-side,  be  joined  with  the  centre  of  the  circle,  the 
sums  of  the  alternate  central  angles  will  be  equal. 

8.  The  sums  of  the  alternate  angles  of  an  encyclic  2n- 
side  are  equal,  namely,  each  sum  is  (n  —  i)  straight  angles. 

9.  The  joins  of  the  ends  of  two  parallel  chords  are  sym- 
metric as  to  the  conjugate  diameter  of  the  chords. 


EXERCISES  III.  139 

10.  A  centre  ray  is  cut  by  two  parallel  tmgents.  Show 
that  the  intercepts  between  tangent  and  circle  are  equal. 

11.  Normals  to  a  chord  from  the  ends  of  a  diameter 
make,  with  the  circle,  equal  intercepts  on  the  chord. 

12.  The  joins  of  the  ends  of  two  diameters  are  parallel  in 
pairs,  and  form  a  rectangle,  and  meet  any  two  parallel  tan- 
gents in  points  symmetric  in  pairs  as  to  the  centre. 

13.  The  joins  of  the  ends  of  two  parallel  chords  meet  the 
tangents  normal  to  the  chords  in  points  whose  other  joins 
are  parallel  to  the  chords. 

14.  A  chord  AB  is  prolonged  to  C,  making  ^C=  radius, 
and  the  centre  ray  CD  is  drawn ;  show  that  one  intercepted 
arc  is  thrice  the  other. 

15.  The  intercepts,  on  a  secant,  of  two  concentric  circles 
are  equal. 

16.  A  chord  through  the  point  of  touch  of  two  tangent 
circles  subtends  equal  central  angles  in  the  circles. 

1 7.  Two  rays  through  the  point  of  touch  of  two  tangent 
circles  intercept  arcs  in  the  circles  whose  chords  are  parallel. 

18.  The  transverse  joins  of  the  ends  of  parallel  diameters 
in  two  tangent  circles  go  through  the  point  of  tangence. 

19.  Four  circles  touch  each  other  outerly  in  pairs:  ist 
and  2d,  2d  and  3d,  3d  and  4th,  4th  and  ist;  show  that  the 
points  of  touch  are  encyclic. 

20.  Show  that  three  circles  drawn  on  three  diameters  OA^ 
OBf  OC  intersect  on  the  sides  of  the  A  ABC. 

21.  Find  the  shortest  and  the  longest  chord  through  a 
point  within  a  circle. 

22.  In  a  convex  4-side  the  sum  of  the  diagonals  is  greater 
than  the  sum  of  two  opposite  sides,  less  than  the  sum  of  all 
the  sides,  and  greater  than  half  the  sum  of  the  sides. 


140  GEOMETRY. 

23.  Three  half-rays  trisect  the  round  angle  O ;  on  each 
is  taken  any  point,  as  A,  B,  C.  Find  a  point  M  such  that 
the  sum  MA  -f  MB  +  MC  is  the  least  possible  (a  minimum). 

24.  Two  tangents  to  a  circle  meet  at  a  point  distant  twice 
the  radius,  from  the  centre  ;  what  angles  do  they  form? 

25.  The  intercept  of  two  circles  on  a  ray  through  one  of 
their  common  points  subtends  a  constant  angle  at  the  other. 

26.  What  is  the  envelope  of  equal  chords  of  a  circle? 

27.  Two  movable  tangents  to  a  circle  intersect  under  con- 
stant angles;  find  the  envelope  of  the  mid-rays  of  these 
angles. 

28.  The  vertex  Fof  a  revolving  right  angle  is  fixed  mid- 
way between  two  parallels,  and  its  arms  cut  the  parallels  at 
A  and  B ;  find  the  envelope  of  AB. 

29.  From  a  fixed  point  P  a  normal  FN  is  drawn  to  a 
movable  tangent  7"  of  a  circle,  and  through  the  mid-point  M 
of /W  there  is  drawn  a  parallel  to  T;  find  its  envelope. 

30.  The  vertices  of  a  A  are  Fj,  Vo,  V^ ;  the  mid-points  of 
its  sides  are  J/j,  M^,  M-^ ;  the  feet  of  its  altitudes  are 
Ai,  A2,  A3 ;  the  inner  bisectors  of  its  angles  meet  the  oppo- 
site sides  at  B^,  B.,,  B^;  and  the  outer  bisectors  at 
B'l,  B'2,  B\  ]  its  centroid  is  C,  its  in-centre  is  /,  its  circum- 
centre  is  S]  its  angles  are  cti,  «2,  «3,  and  their  complements 
are  a\,  a\,  w'g.  Express  through  these  six  angles  the 
angles  between :  (i)  Fi^iandFjFa;  (2)  AiA^Sind  V2V3; 
(3)  AiAoSind  ViAi;  (4)  A1A2  Sind  A^A^;  (5)  M1A2  B.nd 
V,V,;  (6)  J/i^sand  V.V,;  (7)  M,A2SindM,A,;  (8)  A,M2 
and  AiM^;  (9)  A1M2  and  AiA^;  (10)  6^1  and  SV2', 
(11)  SV,  and  V.Vo;  (12)  SF,  and  V2A2;  (13)  /Fi  and 
IF.;   (14)  /Fiand  F^^., ;    (15)    F^'i  and  FoB'.. 


EXERCISES  IV.  141 

31.  Find  the  locus  of  the  mid-points  of  chords  through  a 
fixed  point  upon,  within,  or  without  a  fixed  circle. 

32.  Find  the  locus  of  the  mid-points  of  the  intercepts  of 
a  secant  between  a  fixed  point  and  a  fixed  circle. 

33.  As  the  ends  of  a  ruler  slide  along  two  grooves  normal 
to  each  other,  how  does  its  mid-point  move? 

34.  Two  equal  hoops  move  along  grooves  normal  to  each 
other  and  touch  each  other ;  how  does  the  point  of  touch 
move? 

35.  Orthocentre  O,  centroid  C,  circum-centre  S,  and  cen- 
tre F  oi  Feuerbach's  (9-point)  circle,  of  a  A  are  collinear, 
and  (9C=  2  CS  (Euler),  OF=FS, 

36.  Two  parallel  tangents  meet  two  diameters  of  a  circle 
at  the  vertices  of  a  parallelogram  concentric  with  the  circle. 

37.  The  inner  mid-rays  of  the  angles  of  a  4-side  form  an 
encyclic  4-side. 

38.  The  outer  mid-rays  of  the  angles  of  a  4-side  form  an 
encyclic  4-side.     How  are  the  4-sides  of  37  and  2>^  related? 

39.  The  circum-centres  of  the  four  A  into  which  a  4-side 
is  cut  by  its  diagonals  are  the  vertices  of  a  parallelogram. 

40.  The  circum-centres  of  the  two  pairs  of  A,  into  which 
a  4-side  is  cut  by  its  diagonals  in  turn,  are  how  related  to 
each  other  and  to  the  centres  in  39  ? 

EXERCISES    IV. 

1.  Construct  a  square,  knowing 
{a)  Its  side;  or  (h)  its  diagonal. 

2.  Construct  a  rectangle,  knowing 

(a)  Two  sides;  or  {b)  a  side  and  a  diagonal;  or  (c) 
either  a  side  or  a  diagonal  and  the  angle  of  either  with  the 
other;  or  (//)  a  diagonal  and  its  angles  with  the  other 
diagonal. 


142  GEOMETRY. 

3.  Construct  a  parallelogram,  knowing 

{a)  Two  sides  and  one  angle ;  or  {b)  a  side,  a  diagonal, 
and  the  included  angle ;  or  (<r)  two  sides  and  the  opposite 
diagonal;  or  {d)  two  sides  and  the  included  diagonal;  or 
{e)  two  diagonals  and  a  side;  or  (/)  two  diagonals  and 
their  angles  with  each  other. 

4.  Construct  an  anti-parallelogram,  knowing 

{a)  Its  parallel  sides  and  the  distance  between  them; 
{h)  two  adjacent  sides  and  their  included  angle ;  (r)  two 
adjacent  sides  and  the  angle  between  the  non-parallel  sides ; 
id)  a  diagonal  and  two  adjacent  sides ;  {e)  a  diagonal,  a 
side,  and  the  included  angle. 

5.  Construct  a  kite,  knowing 

{a)  Two  sides  and  an  axis;  {b)  two  sides  and  the  in- 
cluded angle  ;   (r)  .a  side  and  the  axes. 

6.  Construct   the    rays    equidistant    from    three    given 
points. 

7.  Draw  a  ray  through  a  given  point  equally  sloped  to 
two  given  rays. 

8.  A  square  has  one  vertex  at  a  given  point,  and  two 
others  on  two  given  parallel  rays ;  draw  it. 

9.  Hypotenuse  and  sum  of  sides  of  a  right  A  are  given  ; 
draw  it. 

10.  Construct  a  regular  2**-side,  and  a  regular  3.2''-side. 

11.  Find  the  centre  of  a  given  circular  arc. 

12.  Trisect  a  right  angle. 

13.  Two  points,  A  and  B,  of  a  ray  are  given;  find  any 
number  of  points  of  the  ray  without  drawing  it,  and  without 
opening  the  compasses  more  than  AB. 

14.  Find  a  point  on  a  given  ray  or  given  circle  that  has 
a  given  tangent-length  with  respect  to  a  given  circle. 

15.  Through  a  given  point  draw  a  secant  on  which  a 
given  circle  shall  make  a  given  intercept. 


EXERCISES  IV,  143 

1 6.  Draw  four  common  tangents  to  two  given  circles. 

17.  Draw  a  ray  touching  a  given  circle  and  e(iuidistant 
from  two  given  points. 

18.  Draw  a  ray  on  which  two  given  <  ir(  Ics  shall  make 
two  given  intercepts. 

19.  With  three  given  radii  draw  three  circles,  each  touch- 
ing the  other  two. 

20.  Draw  a  circle  toucliing  the  radii  and  the  arc  of  a 
given  sector. 

21.  Draw  a  circle  touching  two  given  equal  intersecting 
circles  and  their  centre  ray. 

22.  On  the  central  intercept  of  two  equal  intersecting 
circles  as  diameter  draw  a  circle  ;  then  draw  a  circle  touch- 
ing the  three  circles. 

23.  Three  equal  circles  touch  each  other  outerly ;  draw 
a  circle  touching  the  three. 

24.  Find  a  point  from  which  two  given  apposed  tracts 
appear  to  be  equal. 

25.  Through  two  given  points  of  a  given  circle  draw  a 
circle  that  shall  cut  a  third  circle  orthogonally. 

26.  Construct  a  A,  knowing 

(rt)  The  feet  of  the  altituOv^s  ;  {b)  the  foot  of  one  altitude 
and  the  mid-points  of  the  other  two  sides ;  {c)  the  three 
ex-centres ;  (^/)  two. ex-centres  and  the  in-centre. 

27.  Draw  through  a  given  point  a  ray  that  shall  form  with 
the  sides  of  a  given  angle  a  A  of  given  perimeter.  Hint : 
Use  the  properties  of  ex-circles. 

28.  Draw  a  5 -side,  knowing  the  mid-points  of  the  sides. 

29.  On  a  tract  AB  there  is  drawn  a  regular  3-side  ;  draw 
on  it  a  regular  6-side.  Generalize  the  problem,  changing 
3  into  «,  6  into  2«,  and  solve  it. 

30.  Given  a  regular  //-side  ;  draw  a  regular  2«-side  having 
liie  original  /;  vertices  for  alternate  vertices.  Do  not  use 
the  circumcircle. 


ELEMENTARY  SYNTHETIC  GEOMETRY 

l)K   TUK 

POINT,   LINE,  AND  CIRCLE  IN  THE  PLANE. 

By  Nathan  F.  Dupuis,  M.A.,  FJtC.S.,  Professor  of  Mathematics  In 
Queen's  College,  Kingston,  Canada.    IGmo.    $1.10. 


FROM  THE  AUTHOR'S  PREFACE. 

"  The  present  work  is  a  result  of  the  author's  exi>erience  in  teaching 
geometry  to  junior  classes  in  the  University  for  a  series  of  years.  It 
is  not  an  edition  of  '  Euclid's  Elements,'  and  has  in  fact  little  relation 
to  that  celebrated  ancient  work  except  in  the  subject-matter. 

"An  endeavor  is  made  to  connect  geometry  with  algebraic  forms 
and  symbols :  (1)  by  an  elementary  study  of  the  modes  of  representative 
geometric  ideas  in  the  symtmls  of  algebra ;  and  (2)  by  determining  the 
consequent  geometric  interpretation  which  is  to  be  given  to  each  inter- 
pretable  algebraic  form.  ...  In  the  earlier  parts  of  the  work  Con- 
structive (Geometry  is  separate<I  from  Descriptive  Greometry,  and  short 
descriptions  are  given  of  the  more  important  geometric  drawing  instru- 
ments, having  special  reference  to  the  geometric  principle  of  their 
actions. . . .  Throughout  the  whole  work  modem  terminology  and 
modern  processes  have  been  used  with  the  greatest  freedom,  regard 
being  had  in  all  cases  to  perspicuity 

"  The  whole  intention  in  preparing  the  work  has  been  to  furnish  the 
student  with  the  kind  of  geometric  knowledge  which  may  enable  him 
to  take  up  most  successfully  the  modern  works  on  analytical  geom- 
etry." 

"  To  this  valuable  work  we  previously  directed  special  attention.  The 
whole  intention  of  the  work  is  to  prepare  the  student  to  take  up  suc- 
cessfully the  modern  works  on  analytical  geometry.  It  is  safe  to  say 
that  a  student  will  learn  more  of  the  science  from  this  book  in  one 
year  than  he  can  learn  from  the  old-fashioned  translations  of  a  certain 
ancient  Greek  treatise  in  two  years.  Every  mathematical  master 
should  study  this  book  in  order  to  learn  the  logical  method  of  present- 
ing the  subject  to  beginners."  —  Canada  Educational  Journal. 


MACMILLAN   &   CO., 

112  FOURTH  AVENUE,    NEW  YORK. 
1 


Mathematical  Works  by  Charles  Smith,  M.A., 

Master  of  Sidney  Sussex  College,  Cambridge. 


A   TREATISE  ON   ALGEBRA.      12mo.     ^1.90.      Key, 

$2.00. 

"Your  Smith's  'Treatise  on  Algebra'  was  vised  in  our  University- 
Classes  last  session,  and  with  very  great  satisfaction.  .  .  .  The  gen- 
eral adoption  of  these  texts  would  mark  an  epoch  in  mathematical 
teaching," — Prof.  W.  B.  Smith,  University  of  Missouri. 

"  Its  style  is  clear  and  neat,  it  gives  alternative  proofs  of  most  of  the 
fundamental  theorems,  and  abounds  in  practical  hints,  among  which 
we  may  notice  those  on  the  resolution  of  expressions  into  factors  and 
the  recognition  of  a  series  as  a  binomial  expansion."  —  Oxford  Review. 

AN  ELEMENTARY  TREATISE  ON  CONIC  SECTIONS. 

New  Edition.    12rao.    $1.60.    Key,  $2.60. 

"  We  can  hardly  recall  any  mathematical  text-book  which  in  neat- 
ness, lucidity,  and  judgment  displayed  alike  in  choice  of  subjects  and 
of  the  methods  of  working,  can  compare  with  this.  .  .  .  We  have  no 
hesitation  in  recommending  it  as  the  book  to  be  put  in  the  hands  of  the 
beginner."  —  Journal  of  Education. 

"  The  best  elementary  work  on  these  curves  that  has  come  under  our 
notice."  —  Academy. 

"A  thoroughly  excellent  elementary  treatise."  —  Nature. 

AN  ELEMENTARY  TREATISE  ON  SOLID  GEOMETRY. 

New  Edition.    12mo.    $2.50. 

' '  The  best  we  can  say  for  this  text-book  is  that  it  is  a  worthy  suc- 
cessor to  the  *  Conies '  previously  noticed  by  us.  .  .  .  Much  credit  is  due 
for  the  freshness  of  exposition  and  the  skill  with  which  the  results  are 
laid  before  the  student."  —  Academy. 

"This  book  is  calculated  to  supply  a  long-felt  want.  The  plan  of 
the  book  is  one  which  will  recommend  itself  to  most.  The  chapter  on 
surfaces  of  the  Second  Degree  falls  rather  earlier  than  is  usual,  but 
the  care  which  has  been  bestowed  upon  it  and  the  clear  explanations 
given  remove  any  difficulties  which  might  otherwise  beset  the  compara- 
tively unprepared  student  in  the  study  of  the  subject.  The  examples 
are  numerous  and  well  selected."  —  Educational  Times. 


MACMILLAN   &  CO., 

112  FOURTH  AVENUE,  NEW  YORK. 
2 


ELEMENTARY   ALGEBRA. 

By  Chaklks  Smith,  M.A.,  Master  of  Sidney  Snasex  College,  Cam- 
bridge. Second  edition,  revised  and  enlargeil.  pp.  viii,  40(.  U'nno. 
5U0. 

FROM  THE  AUTHOR'S  PREFACE. 

"  The  whole  book  has  been  thoroughly  revised,  and  the  early  chapters 
remodelletl  and  Kimplitie<l ;  the  number  of  examples  luis  iM'en  very 
greatly  increased ;  and  chapters  on  Ix)garithms  and  Scales  of  Notation 
have  been  added.  It  is  hoped  that  the  changes  which  have  been  made 
will  increase  the  usefulness  of  the  work." 

From  Prof.  J.  P.  NAYLOR,  of  Indiana  University. 

"  I  consider  it,  without  exception,  tlic  best  Elementary  Alirebra  tliat 
I  have  seen." 

PRESS    NOTICES. 

•*  The  examples  are  numerous,  well  selected,  and  carefully  arranged. 
The  volume  has  many  good  features  in  its  pages,  and  beginners  will 
find  the  subject  thoroughly  placed  before  them,  and  the  road  through 
the  science  made  easy  in  no  small  degree."  —  iSc^oo/ma«rer. 

"There  is  a  logical  clearness  about  his  expositions  and  the  order  of 
his  chapters  for  which  schoolboys  and  schoolmasters  should  be,  and 
will  be,  very  gntetuV* —  Educational  Times. 

"It  is  scientific  in  exposition,  and  is  always  very  precise  and  sound. 
Great  pains  have  been  taken  with  every  detail  of  the  work  by  a  i)erfect 
master  of  the  subject."  —  ^Sc^oo/  Board  Chronicle. 

•'This  Elementary  Algebra  treats  the  subject  up  to  the  binomial 
theorem  for  a  positive  integral  exponent,  and  so  far  as  it  goes  deserves 
the  highest  commendation."  —  Athenaeum. 

**  One  could  hardly  desire  a  better  beginning  on  the  subject  which  it 
treats  than  Mr.  Charles  Smith's  *  Elementary  Algebra.*  .  . .  The  author 
certainly  has  acquired  —  unless  it  •  growed  '  —  the  knack  of  writing 
text-lmoks  which  are  not  only  easily  understood  by  the  junior  student, 
but  which  also  commend  themselves  to  the  admiration  of  more 
matured  oum.**— Saturday  Review. 


MACMILLAN   &  CO., 

112   FOURTH   AVENUE.   NEW  YORK. 
3 


ELEMENTARY  ALGEBRA  FOR  SCHOOLS.     By  H.  S. 

Hall,   M.A.,  and  S.  R.  Kxkhit,  15.A.     Kimo.    Cloth.     Without  an- 
swers, 90  cents.    With  answers,  ^1.10. 

NOTICES  OF  THE   PRESS. 

"...  We  confidently  recommend  it  to  mathematical  teachers,  who, 
we  feel  sure,  will  find  it  the  best  book  of  its  kind  for  teaching  pur- 
poses." —  Nature. 

"  We  will  not  say  that  this  is  the  best  Elementary  Algebra  for  school 
use  that  we  have  come  across,  but  we  can  say  that  we  do  not  remember 
to  have  seen  a  better.  .  .  .  It  is  the  outcome  of  a  long  experience  of 
school  teaching,  and  so  is  a  thoroughly  practical  book.  .  .  .  Buy  or 
borrow  the  book  for  yourselves  and  judge,  or  write  a  better.  ...  A 
higher  text-book  is  on  its  way.  This  occupies  sufficient  ground  for  the 
generality  of  boys."  —  Academy. 

HIGHER  ALGEBRA.  A  Sequel  to  Elementary  Alge- 
bra for  Schools.  By  H.  S.  Hall,  M.A.,  and  S.  R.  Knight,  B.A. 
Fourth  edition,  containing  a  collection  of  three  hundred  Miscella- 
neous Examples  which  will  be  found  useful  for  advanced  students. 
12mo.    $1.90. 

OPINIONS  OF  THE  PRESS. 
**  The  '  Elementary  Algebra '  by  the  same  authors,  which  has  already 
reached  a  sixth  edition,  is  a  work  of  such  exceptional  merit  that  those 
acquainted  with  it  will  form  high  expectations  of  the  sequel  to  it  now 
issued.  Nor  will  they  be  disappointed.  Of  the  authors'  '  Higher  Alge- 
bra,' as  of  their  '  Elementary  Algebra, '  ice  unhesitatingly  assert  that 
it  is  by  far  the  best  work  of  its  kind  loith  which  we  are  acquainted. 
It  supplies  a  want  much  felt  by  teachers."" —  The  Athenseum. 

"  .  .  .  Is  as  admirably  adapted  for  college  students  as  its  predecessor 
was  for  schools.  .  .  .  The  book  is  almost  indispensable  and  will  be 
found  to  improve  upon  acquaintance."  —  The  Academy. 

"...  The  authors  have  certainly  added  to  their  already  high  repu- 
tation as  writers  of  mathematical  text-books  by  the  work  now  under 
notice,  which  is  remarkable  for  clearness,  accuracy,  and  thorough- 
ness. .  .  .  Although  we  have  referred  to  it  on  many  points,  in  no 
single  instance  have  we  found  it  wanting." —  The  School  Guardian. 


MACMILLAN   &  CO., 

112   FOURTH   AVENUE,   NEW  YORK. 
4 


ALGEBRAICAL     EXERCISES     AND     EXAMINATION 

I'AFKKs.  IW  H.  S.  H.vi.L,  .M.A..  hihI  S.  H.  K.mcmt,  B.A.  Kliiio. 
Cloth.     GO  ceiils. 

This  book  has  been  ooinpiled  as  a  suitable  companion  to  the  •'  Ele- 
mentary  Al^rebru  "  b^  the  same  Authors.  It  consists  of  one  hundred 
and  twenty  Projjressjve  Miscellaneous  Exercises  followed  by  a  compre- 
hensive collection  of  Papers  set  at  recent  examinations. 

ARITHMETICAL    EXERCISES    AND    EXAMINATION 

r.\i'KRS.  With  an  ApiHUidix  coutaiuinji  questions  in  L<)Karithn»s 
and  Mensuration.  By  H.  S.  Hall,  M.A.,  and  S.  R.  Knight,  B.A. 
Second  edition.     l(>mo.    (jO  cents. 

KEY  TO  THE  ELEMENTARY  ALGEBRA  FOR  SCHOOLS. 
KEY  TO   THE    HIGHER    ALGEBRA    FOR    SCHOOLS. 

12mo.     .«L'.(K). 

«%  Thcst'  Keys  are  sold  only  upon  a  teacher's  written  order. 

HALL  and  STEVENS.  —  A  Text-Book   of  Euclid's   Ele- 

MENTS.  Including  Alternative  Pr<K)fs,  together  with  additional 
Theorems  and  Exercises,  classified  and  arranged.  By  H.  S.  Hall, 
M.A.,  and  F.  H.  Stkvkns,  M.A. 

Books  I.-VL    Globe  8vo.    31.10. 

Also  sold  separately : 

Book  I.    30  cents.  Books  I.  and  IL    50  cents. 

Books  I.-IV.    75  cents.  Books  III.-VI.    75  cents. 

Book  XI.    30  cents.  Books  V.,  VI.,  XI.  70  cents. 

•'  This  is  a  good  Euclid.  The  text  has  been  carefully  revised,  the  defini- 
tions  simolitiiHl,  and  examples  added  illustrative  of  the  several  propo- 
sitions.   \Ve  predict  for  it  a  very  favorable  reception."  —  Academy. 

LEVETT  and  DAVISON.  — The  Elements  of  Trigonom- 

KTRY.  By  Rawdon  Lkvktt  and  A.  F.  Davison,  Masters  at  King 
Edward's  School,  Birmingham.    Crown  8vo.    $1.U0. 

This  book  is  intended  to  be  a  very  easy  one  for  l)eginners,  all  diffi- 
culties connected  with  the  application  of  algebraic  signs  to  geometry 
anil  with  the  circular  measure  of  angles  l>eing  excluded  from  Part  I. 
Part  II.  deals  with  the  real  algebraical  nuantity,  and  gives  a  fairly 
complete  treatment  and  theory  of  the  circular  and  hyi>erbolic  functions 
con.Hldered  geometrically.  In  Part  III.  complex  numbers  are  dealt  with 
geonaetrically,  and  the  writers  have  tried  to  present  much  of  De  Mor- 
gan's teaching  in  as  simple  a  form  as  possible. 


MACMILLAN   &  CO., 

112  FOURTH  AVENUE,  NEW  YORK. 
6 


ARITHMETIC  FOR  SCHOOLS.     By  Kev.  J.  B.  Lock, 

Fellow  and  Bursar  of  Gonville  and  Caius  College,  Cambridge ;  for- 
merly Master  at  Eton.  Third  Edition,  revised.  Adapted  to  Ameri- 
can Schools  by  Prof.  Charlotte  A.  Scott,  Bryn  Mawr  College, 
Pa.    70  cents. 

The  Academy  says:  "'Arithmetic  for  Schools,' by  the  Rev.  J.  B. 
Lock,  is  one  of  those  works  of  which  we  have  before  noticed  excellent 
examples,  written  by  men  who  have  acquired  their  power  of  present- 
ing mathematical  subjects  in  a  clear  light  to  boys  by  actual  teaching 
in  schools.  Of  all  the  works  which  our  author  has  now  written,  we 
are  inclined  to  think  this  the  best." 

TRIGONOMETRY    FOR    BEGINNERS,    AS    FAR    AS 

the  Solution  of  Triangles.  By  Rev.  J.  B.  Lock.  Third  Edi- 
tion.   75  cents.    Key,  supplied  on  a  teacher's  order  only,  $1.75. 

The  Schoolmaster  says :  "  It  is  exactly  the  book  to  place  in  the  hands 
of  beginners. ,  .  .  Science  teachers  engaged  in  this  particular  branch 
>f  study  will  find  the  book  most  serviceable,  while  it  will  be  equally 
„seful  to  the  private  student." 


TO   BE   PUBLISHED    SHORTLY. 
INTRODUCTORY     MODERN     GEOMETRY     OF     THE 

Point,  Ray,  and  Circle.    By  William  B.  Smith,  Ph.D.,  Profes- 
sor of  Mathematics  in  Missouri  State  University. 

THE  ELEMENTS  OF  GRAPHICAL  STATICS.  A  Text- 
Book  for  Students  of  Engineering.  By  Leander  M.  Hoskins, 
C.E.,  M.S.,  Professor  of  Theoretical  and  Applied  Mechanics,  Univer- 
sity of  Wisconsin. 


Correspondence  from  professors  and  teachers  of  mathematics,  re- 
garding specimen  copies  and  terms  for  introduction,  respectfully 
invited. 


MACMILLAN   &  CO., 

112    FOURTH   AVENUE,   NEW   YORK. 
6 


r 


NIVERSITY  OF  CALIFORNIA  LIBRAiv 
BERKELEY 

Return  to  desk  from  which  borrowed. 
This  book  is  DUE  on  the  last  date  stamped  below. 


MAR  2L'  :m  re 

^EC'D  LD 


LD  21-100m-ll,'49(B7146Bl6)476 


